Ch 08: Conservation of 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

Example #1: Projectile Motion with Energy

Transcript

Hey guys so in this video I want to talk about using the conservation of energy equation to solve projectile motion problems let's check it out. So it says your projectile motion problems asking for speeds or heights are easier to solve using conservation of energy equation, now they're not always easy to solve they're sometimes easy to solve and you can't always do this and I'll talk about this more at the end but we can use energy because speeds have to do with kinetic energy and heights have to do with potential energy so we're going to be able to in some cases use this equation which would be preferable because instead of having to pick one of the three or four equations of motion and have to worry about directions of positive or negative and all that kind of stuff and have to worry about decomposing vectors we can just instead use the energy in some cases to solve it all at once, alright?

So here it says you throw an object 5 kilograms from the top of a 30-meter-tall building so this is 30 meters right here with 20 meters per second and so if you launch with 20 meters per second means that Va right here is 20 meters per second directed at an unknown angle so we don't know this, OK? Above the horizontal it says that the object reaches a maximum height of 40 so if you are at 30 and you get to 40 it's because this additional height here is 10 meters, you could write this or you could just write that the whole thing is 40 ignoring air resistance so there's no work done by friction calculate the object speed at points B C and D, OK? So, let's start over here and I want to find the velocity of object at point B, so I know information about point A I know my initial height at point A and I know my initial velocity at point A and I want to know information about Point B so I'm going to write an energy equation from A to B and it's going to look like this, Ka+Ua+work non-conservative=Kb+Ub, alright? There is kinetic energy at point A because I have a velocity, there's potential at A because I have a height there's no work non-conservative, remember 20 meters per second is after it's already left your hand so while it's between the time that it leaves your hand with 20 and it gets to B over here or actually through the entire motion even you're not doing anything, right? So, the work done by you 0 you did work in throwing it remember but the problem starts counting once it leaves your hand with 20 so the work done by you is 0 and there is no friction there is kinetic energy at point B, alright? Some people sometimes here might make a mistake and think that while the velocity at point B is 0 because the highest point but no the velocity in the Y axis at point B is 0 so the velocity of B is simply made up of its velocity in the X axis which if you remember velocity in the X axis never changes so it's Vax It's whatever the initial velocity in the X axis was, alright? Just as a quick reminder of how some of the stuff in projectile motion works but you didn't really need to know all that for here or remember all of that you just have to not think that the velocity would be 0 at the top, OK Vb is not 0 so I have kinetic energy because Vb is not 0 and I have potential energy because I have a height so I have all four types of energies here and let's do this, so 1/2MVa squared + MGHa = 1/2MVb squared + MGHb. By the way one of the things you could have done here is you could have said to make it a little bit simpler you could have said well since I'm going from A to B I'm going to say that this is a height of 0 and then this is a height of 10 you could have done that if you wanted, alright? I didn't do that here but it doesn't matter it won't affect the final answer it would have just made things a little simpler, here the masses cancel and if I plug in these numbers I have a 1/2(20) squared, gravity I'm going to use 10, the height at point A is 30, Vb is what looking for gravity is 10 and the height at B is 40, alright? You've got all the numbers all you have to do is solve for Vb if you move all the stuff around, OK? I'm going to fast forward a little bit here but if you move all the stuff around and you solve for Vb you get the Vb is 14.1 roughly 14.1 meters per second, OK? So, nothing special here just basic conservation of energy equation, alright? For part B we're being asked what is the velocity at point C, alright? What is Vc? Well to do this we can write an equation from A to C but we actually don't have to do that we don't have to write an equation from A to C at all I should know from conservation of energy that if A and C are at the same height they have to have the same speed, right? If they are at the same height they have to have the same speed so since the height at C is the same as the height at A I can tell you that of the velocity or the speed at C is the same as the speed at A so it's going to be just 20 meters per second we can also think of this in terms of symmetry if you remember symmetry from projectile motion the speed going up is the same as the speed going down for points that are symmetric and because the speed up is the same as the speed down whatever a Vay is the same as Vcy and the velocities in the X axis never change so we Vcx is Vax combine the fact that it's symmetric up and down and that the X axis never changes you end up having a Vc that is exactly the same as Va we talked about this in projectile motion but you can also just have a simple answer to this in terms of energy if the heights are the same the speeds have to be the same, OK? So that's how we get rid of B real quick let me now let me now solve part C. Part C is the velocity at point D, alright? So, to do this we're going to have kinetic initial I'm sorry we're going to write an energy equation from let's say A to D I could have written one from B to D I know information about B to D I can also even have written one from C to D since I know the height of C and the speed of C but it doesn't really matter they all work and they're all equally.... Equal amounts of work. Kd Ud there is a kinetic energy at point A there is a potential energy at point A we're going to say that the ground is 0 over here, there is no work there is kinetic at point D because it's right before you hit the ground and there is no potential energy there so this is a half MVa squared, MGHa, 1/2MVd squared all the masses cancel and we're looking for Vd, let's plug in numbers, 1/2(20)squared + 10 the height is 30 equals 1/2Vd squared if you move everything out of the way this is 200+300, this 2 multiplies over here and then Vd then becomes 31.6, Vd is 31.6 meters per second it should make sense that Vd is greater than Va, right? 20 you go up now you have 14 you come back down to 20 and then you go back down and you're a little bit faster at 31, OK? Vc was 20 and the last question is part what would be the velocity at point D? it says right what is the velocity at point D had it been thrown in any angle below the X axis? So, this is a velocity at point so instead of launching this up you had launched with the Va equals 20 that made some negative angle here in other words you threw it down instead of throwing up, what would it be the velocity at point D? And I can tell you real quick that without having to solve this the velocity would be exactly the same because the direction at which you launch the object doesn't matter because we are doing this using the energy equation, energies are scalars not vectors so the direction does not matter, OK? So I'm going to write here direction does not matter so you have in your notes and you can do this by look at the energy equation and realizing that it just has speed not Vx not Vy not sine of the theta or cosine of theta none of that stuff so there are no thetas, Vx, Vy just V in the energy equation, energy equation, OK? So this means Vd is 31.6 irrespective of the direction that you toss it at even if you didn't straight up it would still have been 31.6 on the way down so that's of that's an important conceptual point you should know that in respect of the direction the final total speed will always be the same because it depends only on the initial speed and the initial height, OK? And that's from conservation of energy, alright I have a second example here let's check that out.

Example #2: Projectile Motion with Energy

Transcript

Alright so in this example we have a ball that we're throwing with 20 meters per second at 37 degrees above the horizontal so letÕs draw this real quick, throwing it this way I'm going to call this Va=20 and it makes an angle of 37 degrees above the horizontal above the X axis like that and it say use the conservation of energy to find its maximum height so let me draw this, it's going to go like this and then hit the ground and I'm going to call this point A, B and C and here I want to know the maximum height so I want to know what is the height at point B? And it's asking us to do this using conservation of energy I know information about Point A I want information about Point B so I'm going to write the energy equation from A to B since A is my known and B is my target so here's the energy equation so there is kinetic energy at point A because you tossed it, it has a speed at that point there's no potential at point A because we're assuming that you're throwing it from the ground or close enough to the ground there's no information about height so we can assume that this is 0, the work done by non-conservative forces is 0 because again you throw a 20 it starts with 20 and then from them instant it leaves your hand with 20 and throughout the entire path on the air you're not doing anything there's no friction there is a kinetic energy at point B even though at point B is the highest point it still has a velocity, it has a velocity at point B is the same as Vx, remember Vx is whatever initial velocity you had here Vx is your Vax and this velocity in the X axis is the same everywhere, Vax is the same as Vx throughout and then in this case there is no Vy here, right? So, I can write that VBy is 0 therefore the entire all of the speed or velocity at point B comes from its X component there is no Y component, OK? So there's definitely energy here and there is energy here because there is a height so let's expand this, 1/2MVa squared = 1/2MVb squared + MGHb, the masses cancel and I'm looking for Hb let's try plugging in some numbers and you're going to notice that we're going to get stuck so 1/2 the speed here is 20, 1/2 and now realize that I don't have Vb at least it wasn't given to me we can calculate Vb but we don't have it right away, Vb square we don't have it + GHb so you got stuck not having a number and this is just the basic physics hustle question as I like to call it because you're going to get stuck and you'll need a number and you're going to have to go and find it, OK? And this is where you have to realize and remember a little bit of projectile motion realize that the velocity at point B because there is no Vby is made up entirely of its X component so I'm going to say that again so since Vby=0 then Vb is made up entirely of Vbx, OK? And that's the first point. Two since the acceleration in the X axis is 0 and it always is 0, right? And unless you have air resistance which most of the time doesn't show up then I can say that Vbx is the same as Vax because it never changes in fact it doesn't change so I like to think of it as not a Vax, Vbx or Vcx but just Vx to make the point to reinforce the point that Vx is always the same so if I want to find Vbx and if I want to find Vbx I just have to find Vax and we can do that Vax is Va cosine of theta X goes with cosine as long as the angle is against the horizontal which it is, OK? And we can plug this in, this is 20 cosine of 37 and 20 cosine of 37 is going to be 16 meters per second so 16 is the number that goes right here now let's continue this over here this is going to be 400, 200 this is 256 divided by this would be 128+Hb if I move this over here I'm going to get a 72 equals 10 Hb so Hb equals 7.2 meters, OK? that's it for this one, again the energy equation got a little weird because we didn't have all the pieces of information and sometimes you're going to get these equation with projectile motion where you are going to have almost everything but then you have to decompose an angle to figure something out and the key thing here to remember was the fact that your Vb at the highest point will always just be your Vx then you can get Vx from Vx so it says here one last point not really related to this question but just to these types of problems that problems that required time notice the neither one of these problems that we've solved require time but that problems require time to be solved cannot be solved with the energy equation and that's because the energy equation has no time it has no reference to time it has no T variable anywhere and instead we're going to have to use motion equations, OK? We're not really going to run into these here because this is from a previous chapter, right? And the whole idea of this portion is that you can do some of these questions using energy but not all of them some of the other questions that you won't be able to do using energy will have time and in that case, you have to use motion equations instead, whenever you're being asked for time use motion equations otherwise you might be able to use the energy equation that's it let me know if you guys have any questions.

Practice: You are practicing jumping as far as you can. In one attempt, you run and leave the floor with 7 m/s directed at an unknown angle. What maximum height do you reach if your speed at that point is 5 m/s? Ignore air resistance.

Practice: When you launch a 3-kg object from the ground with unknown initial speed directed at 37° above the x-axis, it hits the building shown below at 15 m above the ground with 25 m/s. Calculate the object’s launch speed.

Practice: A 3-kg box is nudged off the top of the path shown below, slides down, and is launched form the lower end of the path. The path is frictionless and its highest point is 10 m above the ground. The lower end is 2 m above the ground and makes 53° with the horizontal. Calculate the box’s speed: 

(a) at the lowest point in the path; 

(b) just before it leaves the path; 

(c) at its highest point; 

(d) just before it hits the ground.

An archer performs 20.0 J of work to stretch a bow, storing elastic potential energy in it (much like a spring). The archer then loads an arrow of mass 100 g into the stretched bow, and fires it at an angle of 30.0° above the horizontal. What is the horizontal component of the arrow’s velocity as it leaves the bow A. 20.0 m/s B. 0.63 m/s C. 17.3 m/s D. 10.0 m/s E. 0.55 m/s
A 7 kg projectile is fired at an initial speed of 43 m/s, with a launch angle of 32°. What is the initial kinetic energy of the projectile? What is the kinetic energy of the projectile at its maximum height?
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.10 m and cross the bar with a speed of 0.65 m/s ?
A 0.40-kg ball is thrown with a speed of 8.5 m/s at an upward angle of 36 .What is its speed at its highest point?How high does it go? (Use conservation of energy.)
For its size, the common flea is one of the most accomplished jumpers in the animal world. A 2.10 mm -long, 0.510 mg critter can reach a height of 18.0 cm in a single leap.Neglecting air drag, what is the takeoff speed of such a flea?Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass.If a 71.0 kg , 1.90 m -tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could he jump, and what takeoff speed would he need?In fact, most humans can jump no more than 60.0 cm from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a 71.0 kg person?Where does the flea store the energy that allows it to make such a sudden leap?
A ball is thrown upward with an initial velocity of 15.0 m/s at an angle of 60.0 above the horizontal.Use energy conservation to find the balls greatest height above the ground.
A film of Jesse Owenss famous long jump in the 1936 Olympics shows that his center of mass rose 1.1 m from launch point to the top of the arc. What minimum speed did he need at launch if he was traveling at 6.5 m/s at the top of the arc?
A cannon tilted up at a 31.0 angle fires a cannon ball at 76.0 m/s from atop a 21.0 m -high fortress wall.What is the balls impact speed on the ground below?
In the figure, water balloons are tossed from the roof of a building, all with the same speed but with different launch angles.Which one has the highest speed when it hits the ground? Ignore air resistance.
A 2.8-kg block slides over the smooth, icy hill shown in the figure . The top of the hill is horizontal and 70 m higher than its base.What minimum speed must the block have at the base of the 70-m hill to pass over the pit at the far (righthand) side of that hill?
A sled with rider having a combined mass of 125 kg travels over the perfectly smooth icy hill shown in the accompanying figure.How far does the sled land from the foot of the cliff?
A baseball is thrown from the roof of exttip{h}{h} = 22.1 m -tall building with an initial velocity of magnitude 12.1 m/s and directed at an angle of 53.1 above the horizontal.You may want to review (Page).For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Energy in projectile motion. What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.What is the answer for part (A) if the initial velocity is at an angle of 53.1 below the horizontal?If the effects of air resistance are included, will part (A) or (B) give the higher speed?
A 62-kg skier starts from rest at the top of a ski jump, point A in the figure , and travels down the ramp.If friction and air resistance can be neglected, determine her speed vB when she reaches the horizontal end of the ramp at B.Determine the distance s to where she strikes the ground at C.
It's been a great day of new, frictionless snow. Julie starts at the top of the 60° slope shown in the figure. At the bottom, a circular arc carries her through a 90 m turn, and she then launches off a 3.0-m { m m}high ramp. How far horizontally is her touchdown point from the end of the ramp?
A projectile is fired at an upward angle of 48.0° from the top of a 130-m-high cliff with a speed of 180 m/s m m/s. What will be its speed when it strikes the ground below? (Use conservation of energy. Neglect air resistance.)
A 63 kg skier starts from rest at the top of a ski jump, point A in the figure, and travels down the ramp. If friction and air resistance can be neglected, determine(a) her speed vB when she reaches the horizontal end of the ramp at B(b) the distance s to where she strikes the ground at C
A projectile is fired at an initial speed of v 0 at a launch angle of  θ. Using energy conservation, find the maximum height of the projectile.
A boy starts at rest and slides down a frictionless slide as shown in the figure. The bottom of the track is a height h above the ground. The boy then leaves the track horizontally, striking the ground at a distance d as shown. Using energy methods, determine the initial height H of the boy above the ground in terms of h and d.
A 20.0-kg cannonball is fired from a cannon with muzzle speed of 1 000 m/s at an angle of 37.08 with the horizontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find(a) the maximum height reached by each ball and(b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y =0 at the cannon.
Review. A baseball outfielder throws a 0.150-kg baseball at a speed of 40.0 m/s and an initial angle of 30.0° to the horizontal. What is the kinetic energy of the baseball at the highest point of its trajectory?
A child of mass m starts from rest and slides without friction from a height h along a slide next to a pool (Fig.  P8.27). She is launched from a height h/5 into the air over the pool. We wish to find the maximum height she reaches above the water in her projectile motion.(a) Is the child–Earth system isolated or nonisolated? Why?(b)  Is there a nonconservative force acting within the system?(c) Define the configuration of the system when the child is at the water level as having zero gravitational potential energy. Express the total energy of the system when the child is at the top of the waterslide.(d) Express the total energy of the system when the child is at the launching point.(e)  Express the total energy of the system when the child is at the highest point in her projectile motion.(f) From parts (c) and (d), determine her initial speed vi at the launch point in terms of g and h.(g) From parts (d), (e), and(f), determine her maximum airborne height ymax in terms of h and the launch angle u.(h) Would your answers be the same if the waterslide were not frictionless? Explain.