Ch 07: Work & EnergyWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Concept #1: Calculating Net Work

Transcript

Hey guys, so now that we have seen different ways that we can calculate work in different situations let's talk about how to calculate the total work on an object which is also called the net work of an object or on an object, so the net or total work done on an object is simply the addition of all the works done to the object or on the object, in other words the work done by all forces that act on the object, OK? So work net or net work is simply the sum of all works, so if you have an object and there's three forces let's say acting on it, there are 3 forces acting on the object and they all do work on the object you'd have something like this, Work 1 + Work 2 + Work 3 and if you can find all these numbers or you have all these numbers you just add them up and if you had more obviously you would just keep adding this, OK? But there's another way sometimes you're not going to know all the forces or you're not going all the works but you will know the net force, remember the net force is the sum of all forces this is a vector addition so instead of knowing a bunch of the forces maybe you just know the net force and that works as well, remember work is F D cosine of theta so the work done by the net force is the net work, OK? So, I can do that instead I can just find the net force and multiply it by D cosine of theta, theta is still the angle between your displacement and your net force, OK? So depending on what you have you're going to use one or the other, one point here that's important to make is that work and energy are scalars they're not vectors they don't have direction so unlike forces where we would treat X and Y forces separately because they are vectors with work we combine X and Y to form net work let me give an example real quick if you have a force 3 going this way and a force 4 going this way the net force is not 7 (3+4=7) but instead it's a 5 because it's a vector, now if a force going this way does a work of let's say 10 and a force going this way gives you a total work of 20 then the net work is going to be 30, notice I didn't draw an arrow because work is not a vector, right? When I drew work here is the work done by this force here, OK? So you just add the numbers because they're scalars so let's try this out here's an example, you pull this box with a constant force for a distance of 5 meters up the plane so sort of 5 meters this way, smooth so there's no friction and the magnitude of the work done by each force is shown below so this is the arrow indicates the force not the work, work doesn't have to actually but I'm telling that the work done by normal 0 the work done by F is 75 and the work done by MG is 60, this is the magnitude of the works, OK? And I want to know what is the net work done to each so there's two ways you can calculate net work you can add up all individual works or you can find the net force, here I'm given the work so that's what I'm supposed to do so net work or work net is going to be the sum of all the works which is all these guys added together, the only thing you have to be careful here is that I'm giving the magnitude meaning I'm giving you these works as positives and some of them might be negative and here you have to remember when is a work positive or when does a force do positive work and when does a force do negative work, positive work is in the direction of motion and negative work is against motion, this guy here is 0 because it's perpendicular to motion and so you should remember that the work done by normal is always 0 because it's always perpendicular to motion, this guy goes in the direction of motion so it's +75 and MG is going against the motion because if you remember MGy is perpendicular so it doesn't do anything so the work done by MG is the same as the work done by MGx against the motion. Another thing another way you can remember this is that you were effectively up even though you're going at an incline you're still going up so when you go up gravity does negative work against you, OK? So, when you combine these it's not 75+60 but it's 75+-60 or 75-60 and the answer is 15 Joules, so you do have to realize that this number is a negative very important, OK? And then I'm asking for the net force acting on the box, how do you find net force? Well sort of the straight forward way to find net force is you use this equation the sum of all forces but I don't know any forces right so I'm kind of stuck here and I have really no way of doing this I don't even know the mass of this box so I couldn't even begin to calculate it, so another way that you can find net force now that we've introduced these other equations is by using this right here, right? If you know work net and D cosine of theta you can find F Net, OK? So that's just being able to play with the equations so work net which I now know is 15 is F net, D the distance is 5 cosine of theta, so this is 15 cosine of theta remember when you're trying to figure out the angle there you got to slow down a little bit and be careful, theta is the angle between F Net and D what is the direction of the net force here? Well you should know that the 2 forces that are sort of competing here are this F here that pulls you up and MGX and I know that F has to be greater than MGX because this box is moving up, OK? It's moving up so I know that the direction of the net force has to be going up this way, alright? So, because of that I know that this angle here must be 0, If the net force is up and my delta X is up then this angle is 0, right? So we can now calculate our net force right here, by the way another way that you know this thing is going up is the fact that the work going up is greater than the work going down so you know that more energy is being pumped into moving this thing up than it is being pumped into moving down so because the energy is greater going up this thing will end up going up and therefore I can confidently say that these are in the same direction which means the angle I'm supposed to put there is 0, the cosine of 0 is 1 so I end up with F net=15/5 which is simply 3 Newtons, OK? That's the net force acting on this object, alright? So I want to take a quick point here and then I want to give you guys a practice problem to work on, to find the net work there are two ways one of them is if you have the net force that's great but usually you're going to need to first identify all the forces whether it's because you are going to identify the forces so you can find all the little works and add them up or because you can identify all the forces so you can calculate the net force however you want to do it, OK? But the point is that to find net work you first have to identify forces and the the big idea here is that not all the forces do work so I like to draw this little diagram here that a force may cause a work or a force may do work this means 2 things, one you cannot have work without a force so if there are 5 forces acting on an object there might be 5 work done to the object but either 5 or less because some forces don't some forces don't do work, work is F D cosine of theta so right away you see how you need to have a force in order to have work but remember also that if a force is perpendicular it does no work, the work done by perpendicular forces is 0 so identify all the forces and then figure out which ones do work calculate and add it together, OK? I want you to try this practice problem here and then we're going to keep going so let's give this a shot.

Practice: A 3-kg box is on a flat surface. The box-floor coefficient of friction is 0.6. When you pull horizontally on it for 10 m, it moves with 2 m/s2 . Find the net work on the box. (Start by finding the magnitude of all forces acting on the box)

Practice: A 2-kg box is on a rough horizontal surface. When you pull horizontally on it, it moves with 3 m/s2 . The magnitude of your force and the box-floor coefficient of friction are unknown. What is the net work on the box across 5 m?

Concept #2: The Work-Energy Theorem

Transcript

Hey guys, so in this video I want to talk about the work energy theorem which is basically an equation that's going to connect work with kinetic energy and it's going to allow us to solve problems in a different way let's check it. Alright, so the work energy theorem is this equation right here and it says that the net work or the work net the total work = the change in kinetic energy now change in kinetic energy obviously means K final - K initial, OK? And work net means the sum of all works, alright? So remember we can calculate the work done by a force using this equation and remember that the net work can be calculated using either the sum of each individual work or the work done by the net force you can find the net force by combining all the forces and then find the work done by the net force and remember also that kinetic energy is 1/2MV squared so those are all the questions that we need to talk about the work energy theory so it says here gives us a way to find the initial or final velocity of an object that is acted upon by forces so if there is work done there will be a change in kinetic energy and obviously kinetic energy has a velocity in it so you should you should see a V inside of these guys and these are the kinds of problems we're going to have, you're going to push a box and the moves and we will find some velocities, OK? So, I'm going to do the first example here and I want you guys to try the practice problem.

Example one, a 2-kilogram object has a kinetic energy of 4 at point A and a speed of 3 at point B so I'm giving you two points I don't really give a sense of whether one is to the left or to the right it doesn't really matter but let's start writing this stuff down here, mass=2 I'm going to say that kinetic energy at point A (Ka) is 4 and the speed at point B is 3, OK? Let's just get the information down and then it says find a speed at A, so find a Va notice that I know Ka and then I'm looking for Va well all you have to do is find an equation that relates those two variables and it's pretty easy, the kinetic energy equation ties the two together so not only can I find K if I have V but I can also find V If I have K and that's the idea here, so I'm going to write that Ka=1/2MVa squared, K is given it's a 4, the mass is a 2, so this 2 cancels this 1/2 here and I have the VA is going to be the square root of 4 therefore Va is 2 meters per second, OK? That's it for part B I want to know the kinetic energy at point B so what is Kb? Very similar just backwards I give you Vb and I want to know the kinetic energy at point B, this is even more straightforward Kb is 1/2MV squared at point B as well and we just have to plug in the numbers here, the mass is 2 these two cancel and the velocity is 3 squared so Kb is 9 Joules, OK? That's Va that's Kb 9 Joules right there.

Part C says find the net work from A to B? So, what is the net work from A to B so network I can calculate by doing the sum of all works or by finding the work done by the net force and now there's a third way you can do this and that's the big idea of the work energy theorem is that it gives you a third way to find work net which is by finding the change in kinetic energy, OK? I'm not able to for this problem here I'm not able to find the net work because I don't know all the forces that are acting on this and I don't know the net force so I'm going to have....But I do know velocities and kinetic energies and stuff so I can do it this way ,so if I'm going from A to B. then the network between A to B is just the change in kinetic energy so it's kinetic final or kinetic B- kinetic initial kinetic A and I know these numbers kinetic B we just found it's 9 and kinetic A was given it's a 4 so the net work is 5 Joules, OK? That's it so all we've done is play around with the kinetic energy equation and with the work net equation and the answer is in there somewhere, right? So, cook I want you guys to try practice number two let's give that a shot.

Practice: A 4-kg object has speed 6 m/s at point A, and speed 10 m/s at point B.

(a) How much work was done to it between A and B?
(b) If –32 J of total energy is done to the object between B and C, what speed does it have at C?

Concept #3: The Work-Energy Theorem,ually

Transcript

Hey guys, so now that I've introduced the work energy theorem I want to talk about how that's actually a very important conceptual point, OK? So, it says here that the work energy theorem is also very important conceptually, it's very useful conceptually So it's very useful if you remember that work net is the change in kinetic energy as you're solving physics problems because there's a really important consequence here, well remember kinetic energy is a 1/2MV squared so if your V doesn't change if your speed and not velocity but here it's speed the number, right? So if your V doesn't change if you're V is constant then kinetic energy is constant as well because the mass usually doesn't change so if this number stays the same this number stays the same, so if the V is constant the change in kinetic energy must be 0, there's no change it stays the same and look at the work kinetic energy theorem here, it says that work net is the change in kinetic energy, well if the change kinetic energy is 0 the net work is 0 as well so the idea is that if your speed is constant your net work is 0, OK? Now I said speed not velocity and why is that? Well imagine if you're going round the circle and your speed is always 5 but your velocity is technically changing because your direction changes but on the kinetic energy equation you're plugging in a V here as a 5 so it's a 5 this way and it's still a 5 this way and the kinetic energy is exactly the same if you remember kinetic energy is also a scalar, it doesn't care about direction so it only really depends on speed so if your speed is constant your work net is 0 in fact if you go around a circle with a constant speed uniform circular motion the net work done on you will be 0 and if you remember that you can save a lot of headaches and solve problems much easier, I want to do example 2 and I want you to try practice 2, now practice 2 is a little tricky but I think you have a good chance of getting at least most of it and it's a good sort of mental exercise, alright? So, let's do example 2 here.

I'm going to show you first you know, the answer I'm just going to tell you right now the answer is 0 but I'm going to show you what you might do what you might be doing here wrong if you don't realize that work net is delta K, so it says here a 5 kg box so typical problem solving steps you're going to write your variables on a rough horizontal surface let's draw this here, it's rough so there's a coefficient of friction and your push on it I'm going to draw that as a pull with a constant force, my force is constant cool so that the box moves with a constant speed, so the box is going to move with a constant velocity of 8 meters per second that's constant as well and you might be thinking well how can I have a force and not have an acceleration? Sum of all forces= MA, if I have a force I'm supposed to accelerate, constant speed tells me that the acceleration is 0 how is that possible? Oh it's because there has to be a friction which in case of box kinetic friction that's exactly canceling with my force, so in a regular problem solving....In a regular steps of solving this problem you might have thought about all of this stuff and now if you want to find the net work maybe you're thinking the net work here is the work done by F + the work done by kinetic friction obviously MG does no work and normal does no work but look at how long that analysis is and if you were to calculate all the stuff you would've found in fact that the answer is 0 and that's because whatever you have here will be a positive and that number will be the same number here as a negative or you could have just seen that it says constant velocity or constant speed and then right away you would have known that the answer is 0 and you wouldn't have to do any of this crap, OK? So, I actually want you to do this I want you to scratch this out so when you're reviewing this you remember that you don't have to go to all this, OK? So, the answer would have just been 0 so if you get this on a test it most likely will be multiple choice for those of you that have that and then you'll be able to kill right away, alright? So, I want you to practice number two let's give that a shot.

Practice: You lift a 3 kg object from the floor to a height of 2 m. Find the:

(a) work done by you;
(b) work done by gravity;
(c) net work done on the object. You then walk horizontally with the object for 10 m.
(d) How much work do you do?

A mass of 1kg begins at a position (–3 m, –4 m) at a speed of 4.6 m/s. If it moves in a straight line and ends up at a position (2.5 m, 3 m) at a speed of 3.2 m/s, how much work was done on the mass? What is the net force acting on the mass?
An 8.1 kg crate is pulled 5.1 m up a 30 incline by a rope angled 18 N above the incline. The tension in the rope is 130 (Pages 222 - 223) and the crates coefficient of kinetic friction on the incline is 0.25.You may want to review (Pages 222 - 223) . For help with math skills, you may want to review: The Vector Dot Product How much work is done by tension, by gravity, and by the normal force?What is the increase in thermal energy of the crate and incline?
A 16-kg sled starts up a 28 incline with a speed of 2.3 m/s . The coefficient of kinetic friction is exttip{mu_{ m k}}{mu_k} = 0.26.How far up the incline does the sled travel?What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part (a)?If the sled slides back down, what is its speed when it returns to its starting point?
Proper design of automobile braking systems must account for heat buildup under heavy braking.Calculate the thermal energy dissipated from brakes in a 1500 kg car that descends a 18 hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.31 km measured along the road.
You are a member of an Alpine Rescue Team and must project a box of supplies up an incline of constant slope angle exttip{alpha }{alpha} so that it reaches a stranded skier who is a vertical distance exttip{h}{h} above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient exttip{mu _{ m k}}{mu_k}.You may want to review (Page). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Using work and energy to calculate speed.Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of exttip{g}{g}, exttip{h}{h}, exttip{mu _{ m k}}{mu_k}, and exttip{alpha }{alpha}.
An old oaken bucket of mass 6.75 kg hangs in a well at the end of a rope. The rope passes over a frictionless pulley at the top of the well, and you pull horizontally on the end of the rope to raise the bucket slowly a distance of 4.00 m.How much work do you do on the bucket in pulling it up?How much work is done by the gravitational force acting on the bucket?What is the total work done on the bucket?
A factory worker pushes a crate of mass 30.1 kg a distance of 4.55 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.What magnitude of force must the worker apply?How much work is done on the crate by this force?How much work is done on the crate by friction?How much work is done by the normal force?How much work is done by gravity?What is the total work done on the crate?
A factory worker pushes a 30.0 kg crate a distance of 3.3 m along a level floor at constant velocity by pushing downward at an angle of 30 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.What magnitude of force must the worker apply to move the crate at constant velocity?How much work is done on the crate by this force when the crate is pushed a distance of 3.3 m ?How much work is done on the crate by friction during this displacement?How much work is done by the normal force?How much work is done by gravity?What is the total work done on the crate?
Two blocks are connected by a very light string passing over a massless and frictionless pulley (the figure ). Traveling at constant speed, the 20.0-N block moves 76.0 cm to the right and the 12.0-N block moves 76.0 cm downward.During this process, how much work is done on the 12.0-N block by the tension in the string?During this process, how much work is done on the 20.0-N block by the normal force?Find the total work done on 12.0-N block.During this process, how much work is done on the 12.0-N block by gravity?During this process, how much work is done on the 20.0-N block by gravity?During this process, how much work is done on the 20.0-N block by the tension in the string?During this process, how much work is done on the 20.0-N block by friction?Find the total work done on 20.0-N block.
You push your physics book 1.80 m along a horizontal tabletop with a horizontal push of 2.60 N while the opposing force of friction is 0.650 N . You may want to review (Pages 173 - 177). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Work done by a constant force.How much work does your 2.60 N push do on the book?How much work does the friction force do on the book?How much work does the normal force from the table do on the book?How much work does gravity do on the book?What is the net work done on the book?
A batter hits a baseball with mass 0.150 kg straight upward with an initial speed of 28.0 m/sHow much work has gravity done on the baseball when it reaches a height of 26.0 m above the bat?Use the work-energy theorem to calculate the speed of the baseball at a height of 26.0 m above the bat. You can ignore air resistance.Does the answer to part B depend on whether the baseball is moving upward or downward at a height of 26.0 m ?Explain.
Use the work-energy theorem to solve each of these problems. You can use Newtons laws to check your answers. Neglect air resistance in all cases.A branch falls from the top of a 95.7-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground?A volcano ejects a boulder directly upward 526 m into the air. How fast was the boulder moving just as it left the volcano?
You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.1 m above the ground, it is traveling at 25.1 m/s upward.Use the work-energy theorem to find the rocks speed just as it left the ground.Use the work-energy theorem to find its maximum height.
In the Bohr model of the atom, the ground-state electron in hydrogen has an orbital speed of 2190 km/s.The mass of an electron is 9.1110-31. What is its kinetic energy?If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 m, how many joules of kinetic energy will it have when it reaches the ground?Is it reasonable that a 30-kg child could run fast enough to have 100 J of kinetic energy?
A cable with 21.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. You may want to review (Pages 251 - 252).What is the blocks speed after being lifted 2.00 m? Solve this problem using work and energy.
A pile driver lifts a 350 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground. A fall of 1.5 m before striking the pipe drives the pipe in 45 cm.What is the average force exerted on the pipe?
The gravitational attraction between two objects with masses m1 and m2, separated by distance x, is F=Gm1m2/x2, where G is the gravitational constant.How much work is done by gravity when the separation changes from x1 to x2? Assume x2<x1.If one mass is much greater than the other, the larger mass stays essentially at rest while the smaller mass moves toward it. Suppose a 1.5kg1013 kg comet is passing the orbit of Mars, heading straight for the 1.99kg1030 kg sun at a speed of 3.7 m/s104 G. What will its speed be when it crosses the orbit of Mercury? The orbit of Mars is 2.28m/s108 m km, the orbit of Mercury is 5.79km107 m km, and G = 6.67km10-11 m Ncdot m^2/kg^2.
Youre fishing from a tall pier and have just caught a 1.3 kg fish. As it breaks the surface, essentially at rest, the tension in the vertical fishing line is 16 N.Use work and energy to find the fishs upward speed after youve lifted it 2.0 m?
A ball shot straight up with kinetic energy K0 reaches height h.What height will it reach if the initial kinetic energy is doubled?
You throw a 5.5 g coin straight down at 4.0 m/s from a 55-m-high bridge.How much work does gravity do as the coin falls to the water below?What is the speed of the coin just as it hits the water?
The figure shows how the force exerted by the string of a compound bow on an arrow varies as a function of how far back the arrow is pulled (the draw length). Assume that the same force is exerted on the arrow as it moves forward after being released. Full draw for this bow is at a draw length of 75.0 cm.If the bow shoots a 0.0250-kg arrow from full draw, what is the speed of the arrow as it leaves the bow?
The mass of a proton is 1836 times the mass of an electron.A proton is traveling at speed V . At what speed (in terms of V ) would an electron have the same kinetic energy as the proton?An electron has kinetic energy K . If a proton has the same speed as the electron, what is its kinetic energy (in terms of K )?
On April 13, 2029 (Friday the 13th!), the asteroid 99942 Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver? Take the speed of impact with the earth to be 12.6 km/s. Neglect the work that would be done by gravity.The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184 1015 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
T. Rex. The dinosaur Tyrannosaurus rex is thought to have had a mass of about 7000 kg.Treat the dinosaur as a particle and estimate its kinetic energy at a walking speed of 4 km/h.With what speed would a 70.0-kg person have to move to have the same kinetic energy as a walking T. rex?
At room temperature, an oxygen molecule, with mass of 5.31 10 - 26 kg, typically has a kinetic energy of about 6.21 10 - 21 J.How fast is it moving?
Which has the larger kinetic energy, a 10 g bullet fired at 500 m/s or a 75 kg bowling ball rolled at 5.5 m/s ?
If the kinetic energy of a particle is tripled, by what factor has its speed increased?If the speed of a particle is halved, by what factor does its kinetic energy change?
At what speed does a 1600 kg compact car have the same kinetic energy as a 1.80×104 kg truck going 24.0 km/hr ?
A mother has four times the mass of her young son. Both are running with the same kinetic energy.What is the ratio vs/vm other of their speeds?
One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 7.5 m/s , they then have the same kinetic energy.What were the original speeds of the two cars?
A 3.0 g locust reaches a speed of 3.0 m/s during its jump.What is its kinetic energy at this speed?If the locust transforms energy with 40 % efficiency, how much energy is required for the jump?
A 33.0 kg crate is initially moving with a velocity that has magnitude 4.01 m/s in a direction 37.0 west of north.How much work must be done on the crate to change its velocity to 6.02 m/s in a direction 63.0 south of east?
If a particles speed increases by a factor of 5, by what factor does its kinetic energy change?
Adult cheetahs, the fastest of the great cats, have a mass of about 70.0 kg and have been clocked at up to 72.0 mph (32.2 m/s ).You may want to review (Pages 177 - 182). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Using work and energy to calculate speed.How many joules of kinetic energy does such a swift cheetah have?By what factor would its kinetic energy change if its speed were doubled?
About 50000 years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Recent (2005) measurements estimate that this meteor had a mass of about 1.4 108 kg (around 150000 tons) and hit the ground at 12 km/s.You may want to review (Pages 177 - 182). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Using work and energy to calculate speed.How much kinetic energy did this meteor deliver to the ground?How does this energy compare to the energy released by a 1.0-megaton nuclear bomb? (A megaton bomb releases the same energy as a million tons of TNT, and 1.0 ton of TNT releases 4.184 109 J of energy.)
A moving electron has kinetic energy K1 . After a net amount of work W has been done on it, the electron is moving one-quarter as fast in the opposite direction.Find W in terms of K1 .Does your answer depend on the final direction of the electrons motion?
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N.How much work is done by tension?
A 53.0 kg ice skater is gliding along the ice, heading due north at 4.20 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but exttip{mu _{ m k}}{mu_k} =0. Suddenly, a wind from the northeast exerts a force of 4.10 N on the skater.Use work and energy to find the skaters speed after gliding 100 m in this wind.What is the minimum value of exttip{mu _{ m s}}{mu_s} that allows her to continue moving straight north?
A sled with mass 12.10 kg moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4.51 m/s ; after it has traveled a distance 2.51 m beyond this point, its speed is 6.61 m/s .Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sleds motion.
A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface . The block is originally revolving at a distance of 0.41 m from the hole with a speed of 0.71 m/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.15 m . At this new distance, the speed of the block is 1.94 m/s .What is the tension in the cord in the original situation when the block has speed exttip{v_{ m 0}}{v_0} = 0.71 m/s ?What is the tension in the cord in the final situation when the block has speed exttip{v_{ m 1}}{v_1} = 1.94 m/s ?How much work was done by the person who pulled on the cord?
A 0.2 kg plastic cart and a 20 kg lead cart both roll without friction on a horizontal surface. Equal forces are used to push both carts forward a distance of 1 m, starting from rest.After traveling 1 m, is the kinetic energy of the plastic cart greater than, less than, or equal to the kinetic energy of the lead cart?
One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 4.00-kg can of beans is attached to the other end. The spring is initially neither stretched nor compressed. A constant horizontal force of 56.0 N is then applied to the can, in the direction away from the post.What is the speed of the can when the spring is stretched 0.400 m?At the instant the spring is stretched 0.400 m, what is the magnitude of the acceleration of the block?At the instant the spring is stretched 0.400 m, what is the magnitude of the acceleration of the block?
A block of ice of mass 4.10 kg is placed against a horizontal spring that has force constant exttip{k}{k} = 210 N/m and is compressed a distance 2.60×10−2 m . The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring.Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length.What is the speed of the block after it leaves the spring?
A 45 g rock is placed in a slingshot and the rubber band is stretched. The magnitude of the force of the rubber band on the rock is shown by the graph in . The rubber band is stretched 30 cm and then released.What is the speed of the rock?
You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1300 kg car moving at 0.66 m/s is to compress the spring no more than 7.6×10−2 m before stopping.What should be the force constant of the spring? Assume that the spring has negligible mass.
A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 6.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring?How fast was the box going when it hit the spring?
Chris jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle as shown in the figure . He falls for 15 m before the bungee cord begins to stretch. Chriss mass is 75 kg and we assume the cord obeys Hookes law, F = - k x, with exttip{k}{k} = 55 N/m .If we neglect air resistance, estimate how far below the bridge Chriss foot will be before coming to a stop. Ignore the mass of the cord (not realistic, however) and treat Chris as a particle.
A 6.0-kg box moving at 5.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. You may want to review (Pages 183 - 189). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Motion with a varying force.Use the work-energy theorem to find the maximum compression of the spring.
At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant exttip{k}{k} = 4100 N/m and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 71.0 kg are pushed against the other end, compressing the spring 0.380 m . The sled is then released with zero initial velocity.You may want to review (Pages 183 - 189). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Motion with a varying force.What is the sleds speed when the spring returns to its uncompressed length?What is the sleds speed when the spring is still compressed 0.205 m ?
A car is stopped in a distance exttip{D}{D} by a constant friction force that is independent of the cars speed.What is the stopping distance (in terms of exttip{D}{D}) if the cars initial speed is tripled?What is the stopping distance (in terms of exttip{D}{D}) if the speed is the same as it originally was but the friction force is tripled? (Solve using the work-energy theorem.)
At an accident scene on a level road, investigators measure a cars skid mark to be 93 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.Why does the cars mass not matter?
A train is moving along a track with constant speed v1 relative to the ground. A person on the train holds a ball of mass m and throws it toward the front of the train with a speed v2 relative to the train.Calculate the change in kinetic energy of the ball in the Earth frame of reference.Calculate the change in kinetic energy of the ball in the train frame of reference.How much work was done on the ball in the Earth frame of reference?How much work was done on the ball in the train frame of reference?
A 1.50 kg book is sliding along a rough horizontal surface. At point exttip{A}{A} it is moving at 3.21 m/s , and at point exttip{B}{B} it has slowed to 1.25 m/s .How much work was done on the book between exttip{A}{A} and exttip{B}{B}?If -0.750J of work is done on the book from exttip{B}{B} to exttip{C}{C}, how fast is it moving at point exttip{C}{C}?How fast would it be moving at exttip{C}{C} if 0.750J of work were done on it from exttip{B}{B} to exttip{C}{C}?
On an essentially frictionless, horizontal ice rink, a skater moving at 6.0 m/s encounters a rough patch that reduces her speed by 46 % due to a friction force that is 26 % of her weight.Use the work-energy theorem to find the length of this rough patch.
A 65 kg softball player slides into second base, generating 750 J of thermal energy in her legs and the ground.How fast was she running?
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 m by a trained dog who exerts a horizontal force with magnitude 36.0 N. You may want to review (Pages 177 - 182). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Using work and energy to calculate speed.Use the work-energy theorem to find the final speed of the 12-pack if there is no friction between the 12-pack and the floor.Use the work-energy theorem to find the final speed of the 12-pack if the coefficient of kinetic friction between the 12-pack and the floor is 0.30.
A little red wagon with mass 7.10 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 3.60 m/s and then is pushed 4.1 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. You may want to review (Pages 177 - 182). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Using work and energy to calculate speed.Use the work-energy theorem to calculate the wagons final speed.Use this acceleration in the kinematic relationship vx2 = v0x2 + 2ax (x - x0 ) to calculate the wagons final speed.Calculate the acceleration produced by the force.
How much work is required to stop an electron ( m = 9.11 10 - 31 kg ) which is moving with a speed of 1.60×106 m/s ?
A car is traveling on a level road with speed v0 at the instant when the brakes lock, so that the tires slide rather than roll.You may want to review (Pages 177 - 182). For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Using work and energy to calculate speed.Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of v0, g, and the coefficient of kinetic friction k between the tires and the road.By what factor would the minimum stopping distance change if the coefficient of kinetic friction were doubled ?By what factor would the minimum stopping distance change if the initial speed were doubled?By what factor would the minimum stopping distance change if both the coefficient of kinetic friction and the initial speed were doubled?
Your cat "Ms." (mass 7.00 kg) is trying to make it to the top of a frictionless ramp 2.00 m long and inclined upward at 30.0 above the horizontal. Since the poor cat cant get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 100 { m N} force parallel to the ramp.If Ms. takes a running start so that she is moving at 2.40 m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.
How much work is done by the horizontal force FP = 150 N on the 18-kg block of the figure when the force pushes the block 6.0 m up along the 32 frictionless incline?How much work is done by the gravitational force on the block during this displacement?How much work is done by the normal force?What is the speed of the block (assume that it is zero initially) after this d is placement? [Hint: Work-energy involves net work done.]
Starting from rest, a crate of mass m is pushed up a frictionless slope of angle heta by a horizontal force of magnitude F. Use work and energy to find an expression for the crates speed v when it is at height h above the bottom of the slope.Doug uses a 28 N horizontal force to push a 4.1 kg crate up a 1.8 m -high, 20 frictionless slope. What is the speed of the crate at the top of the slope?
A block of ice with mass 2.00 kg slides 1.61 m down an inclined plane that slopes downward at an angle of 36.9 below the horizontal.If the block of ice starts from rest, what is its final speed? Ignore friction.
A box of mass 5.5 kg is accelerated from rest by a force across a floor at a rate of 2.6 m/s2 for 5.6 s .Find the net work done on the box.
An 200g object initially at rest slides 50 cm down a frictionless incline at a 40° angle. At the bottom of the slope, the object encounters a patch of rough surface, which causes it to come to a stop after 1 m. What is the coefficient of kinetic friction of this rough patch?
How much work must be done to stop a 1300 kg car traveling at 90 km/h?
A 145 g baseball traveling 33 m/s moves a fielders glove backward 25 cm when the ball is caught. What was the average force exerted by the ball on the glove?
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 240 N . For the first 11.5 m the floor is frictionless, and for the next 11.5 m the coefficient of friction is 0.19. What is the final speed of the crate after being pulled these 23.0 m?
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30 degrees above the floor. The tension is a constant 31.0 N { m N}and the coefficient of friction is 0.200. Use work and energy to find Paul's speed after being pulled 3.10 m.
A 6.10 kg block is pushed 9.30 m up a smooth 38.0  inclined plane by a horizontal force of 76.0 N . If the initial speed of the block is 3.30 m/s up the plane, calculate the:(a) initial kinetic energy of the block(b) work done by the 76.0 N force(c) work done by gravity(d) work done by the normal force(e) final kinetic energy of the block
A soccer ball with mass 0.460 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 41.0 N in the same direction as the balls motion. Over what distance must the players foot be in contact with the ball to increase the balls speed to 6.00 m/s?
A block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless, horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle θ = 25.0° below the horizontal as shown in the figure. Determine the work done on the block by(a) the applied force(b) the normal force exerted by the table(c) the gravitational force(d) the net force on the block
A 500 kg elevator accelerates upward at 1.3 m/s2 for 19 m, starting from rest.(a) How much work does gravity do on the elevator?(b) How much work does the tension in the elevator cable do on the elevator?(c) What is the elevator’s kinetic energy after traveling 19 m?
The figure shows the kinetic-energy graph for a 2.0 kg object moving along the x-axis. Determine the work done on the object during each of the four intervals AB, BC, CD, and DE.
A 400 g particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 9.0 m/s at x = 0 m. What is its velocity at x = 3 m?
A box with mass 5.00 kg is pulled up a 36.9° incline by a constant force  F that has magnitude 75.0 N and that is parallel to the incline. The distance along the incline from the bottom to the top is 6.00 m. During the motion of the box, the surface of the incline exerts a constant friction force fk = 18.0 N on the box, in a direction opposite to the motion.If the box starts froom rest at the botttom of the incline, what is the kinetic energy of the box when it reaches the top of the incline?
A 2.30-kg textbook is forced against a horizontal spring of negligible mass and force constant 260 N/m , compressing the spring a distance of 0.260 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction μk=0.30. Use the work-energy theorem to find how far the textbook moves from its initial position before coming to rest.
A horizontal spring with spring constant 270 N/m is compressed by 20 cm and then used to launch a 300 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?
A 10 kg object slides down an incline of 30 o. If the object starts at a height of 30 cm, and feels a coefficient of kinetic friction of 0.2 down the slope, answer the following questions:a. How much work is done on the object as it moves down the slope?   b. What is the change in the object’s potential energy down the slope?    c. What is the change in the object’s kinetic energy down the slope
Particle A has half the mass and eight times the kinetic energy of particle B. What is the speed ratio vA / vB?
If the human body could convert a candy bar directly into work, how high could a 76 kg man climb a ladder if he were fueled by one bar (=1100 kJ)?If the man then jumped off the ladder, what will be his speed when he reaches the bottom?
Use the work–energy theorem to solve each of these problems.(a) A skier moving at 5.01  m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?(b) Suppose the rough patch in part A was only 2.91  m long. How fast would the skier be moving when she reached the end of the patch?(c) At the base of a frictionless icy hill that rises at 25.0 above the horizontal, a toboggan has a speed of 12.1 { m ; m/s} toward the hill. How high vertically above the base will it go before stopping?
A truck has four times the mass of a car and is moving with twice the speed of the car. If Kt and Kc refer to the kinetic energies of truck and car respectively, it is correct to say that A) Kt = 16 Kc. B) Kt = 4 Kc. C) Kt = 2 Kc. D) Kt = Kc. E) Kt = 1/2 Kc.
A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0o with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m.(a) How much work is done by the gravitational force on the crate?(b) Determine the increase in internal energy of the crate–incline system owing to friction.(c) How much work is done by the 100-N force on the crate?(d) What is the change in kinetic energy of the crate?(e) What is the speed of the crate after being pulled 5.00 m?
Jonathan is riding a bicycle and encounters a hill of height 7.30 m. At the base of the hill, he is traveling at 6.00 m/s. When he reaches the top of the hill, he is traveling at 1.00 m/s. Jonathan and his bicycle together have a mass of 85.0 kg. Ignore friction in the bicycle mechanism and between the bicycle tires and the road.(a) What is the total external work done on the system of Jonathan and the bicycle between the time he starts up the hill and the time he reaches the top?(b) What is the change in potential energy stored in Jonathan’s body during this process?(c) How much work does Jonathan do on the bicycle pedals within the Jonathan–bicycle–Earth system during this process?
Jonathan is riding a bicycle and encounters a hill of height h. At the base of the hill, he is traveling at a speed vi . When he reaches the top of the hill, he is traveling at a speed vf . Jonathan and his bicycle together have a mass m. Ignore friction in the bicycle mechanism and between the bicycle tires and the road. (a) What is the total external work done on the system of Jonathan and the bicycle between the time he starts up the hill and the time he reaches the top? (b) What is the change in potential energy stored in Jonathan’s body during this process? (c) How much work does Jonathan do on the bicycle pedals within the Jonathan– bicycle–Earth system during this process?
An 80.0-kg skydiver jumps out of a balloon at an altitude of 1 000 m and opens his parachute at an altitude of 200 m.(a) Assuming the total retarding force on the skydiver is constant at 50.0 N with the parachute closed and constant at 3 600 N with the parachute open, find the speed of the skydiver when he lands on the ground.(b) Do you think the skydiver will be injured? Explain.(c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s?(d) How realistic is the assumption that the total retarding force is constant? Explain.
An older-model car accelerates from 0 to speed v in a time interval of Δt. A newer, more powerful sports car accelerates from 0 to 2v in the same time period. Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars.
A 0.600-kg particle has a speed of 2.00 m/s at point Ⓐ and kinetic energy of 7.50 J at point Ⓑ. What is (a) its kinetic energy at Ⓐ, (b) its speed at Ⓑ, and (c) the net work done on the particle by external forces as it moves from Ⓐ to Ⓑ?
Review. You can think of the work–kinetic energy theorem as a second theory of motion, parallel to Newton’s laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a), (b), and (c) separately from parts (d) and (e) so you can compare the predictions of the two theories. A 15.0-g bullet is accelerated from rest to a speed of 780 m/s in a rifle barrel of length 72.0 cm. (a) Find the kinetic energy of the bullet as it leaves the barrel. (b) Use the work–kinetic energy theorem to find the net work that is done on the bullet. (c) Use your result to part (b) to find the magnitude of the average net force that acted on the bullet while it was in the barrel. (d) Now model the bullet as a particle under constant acceleration. Find the constant acceleration of a bullet that starts from rest and gains a speed of 780 m/s over a distance of 72.0 cm. (e) Modeling the bullet as a particle under a net force, find the net force that acted on it during its acceleration. (f) What conclusion can you draw from comparing your results of parts (c) and (e)?
A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. The coefficient of friction between box and floor is 0.300. Find (a) the work done by the applied force, (b) the increase in internal energy in the box–floor system as a result of friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box.
Review. A 7.80-g bullet moving at 575 m/s strikes the hand of a superhero, causing the hand to move 5.50 cm in the direction of the bullet’s velocity before stopping. (a) Use work and energy considerations to find the average force that stops the bullet. (b) Assuming the force is constant, determine how much time elapses between the moment the bullet strikes the hand and the moment it stops moving.