Practice: You pull a 5-kg object vertically up with a constant 100-N for 2 m. How much work do you do?

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Intro to Energy | 15 mins | 0 completed | Learn |

Intro to Calculating Work | 36 mins | 0 completed | Learn |

Work By Gravity & Inclined Planes | 40 mins | 0 completed | Learn |

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Concept #1: Work Done by a Constant Force

**Transcript**

Hey guys, so in this video we're going to start talking about work as well as how to calculate that quantity Let's check it out. So, the idea is that when you push against a box on a smooth surface so let's draw this as a smooth surface there is no friction I can say mu=0 if you push it letÕs say this way the box starts to move, right? There's no resistance if you push it has to start moving F=MA you get acceleration it starts accelerating it moves, since it begins to move it now has a V and if it has a V it gains kinetic energy, OK? And if you remember kinetic energy is 1/2MV squared so if I have a V I have a kinetic energy, well what does that energy come from? That energy comes from you from your push, OK? From F over here so that energy comes from you, you ate some food and the energy ultimately comes from the sun, right? The sun gives energy to the plant, the cow eats the plant and you eat the cow and then you push this box, right? Very simplistic but that's basically the idea, this transfer of energy you transferred your stored energy into the box into the form of kinetic energy, this transfer of energy from you to the box it's called work. OK? So, in physics that's what works means and we say that you do work on the box, OK? So, do you work on an object is transferring energy into that objects, OK? The amount of work that a force does an object is the amount of energy that it gives to that object or takes away from that object so if you push a box and you make it faster you're giving it energy and therefore we're going to say we'll talk more about this later but we're going to see you do positive work if a box is moving and you push on it so that it slows down you're stealing kinetic energy from that box so you're taking energy away from the box so you do negative work we'll talk more about positive or negative work later but I want you to understand that the idea of work is that work is energy and transfer it's you transferring energy, OK? To calculate work the equation for the work done by a constant force this only works for constant forces can be calculated using this equation, OK? I'm going to give you sort of the fancy definition than the sort of useful one, so it's force which is a vector the magnitude of force times the magnitude of displacement but these guys are supposed to be parallel to each other, OK? So that means that they're supposed to be parallel to each other and the other less fancy version of this would be F delta X cosine of theta if they're not parallel to each other then you would do this F delta X cosine of theta where theta (I'll write this here in a second) is the angle between F and delta X, these two are vectors you can draw two little arrows and this is just the angle between them, one way that I like to write this equation is I put a little arrow here to remind me that theta is the angle between the two guys that are in the equation hanging out next to theta. Now this is a lot of lines and arrows everywhere so the simplest the simpler version of this will be F D cosine of theta, OK? Now that being said remember this was originally absolute value so when you plug these make sure the F is positive and D is positive, OK? Even if you're moving to the left or the forces is to the left it doesn't matter these guys will always be positive and this is the angle, cool? Now last point here since work is energy transferred work is just transfer of energy the unit of work is the same as a unit of energy Joules, OK? Let's do some problems real quick I'll do one and I want you guys to do one.

So, a 2 Kg box is pulled horizontally by a 3 Newton force let's do that real quick, 2-kilogram box is pulled the horizontally by a force of 3 Newtons for 5 meters that's my distance, so I did this for a distance of 5 meters so the idea is I pull and once the box has covered 5 meters I just let it go, right? I want to know what is the work done by this force, OK? So, works are done by forces, the work done by a force F is right here F D cosine of theta, OK? Now F and D are pretty straight forward the force is 3 D is 5 but every time you get to theta I need you to slow down and take it easy here and be careful because there is some question that are going to be tricky remember it's the angle between these two guys, forces is this way displacement is this way notice how they are parallel to each other they go in the same direction, the angle when two things are parallel to each other is 0 degrees so this is cosine of 0 and the cosine of 0 you plug into the calculator and you see that it's one so the final answer here is just 3x5x1 or 15 Joules...15 Joules, that's your work. Alright so I now want you to try the second problem it's in a vertical axis but I think you can figure this out let's give this a shot.

Practice: You pull a 5-kg object vertically up with a constant 100-N for 2 m. How much work do you do?

Example #1: Work by a Constant Force

**Transcript**

Alright, so here another example I have a 10 kilograms box on a smooth surface so no friction, it's pulled for a distance of 20 so this is my D or delta X as shown below find the work done by F, letÕs find a work done by F. So, this is very straightforward the work done by F is F D cosine of theta, the force is 50 the distance is 20 and then the cosine of the angle between the two, OK? Because this angle is the angle between these two guys I don't really have to decompose F, remember one of the first things you should do and you should just think of this instinctively almost when you see a force at an angle is to decompose it, here we don't have to because I can just put the angle between them here, OK? Remember it's the angle between force which is this way if you pull this way remember whenever you pull up at an angle the object still going to move this way otherwise would come off the floor that would be weird so the angle you should do is the angle between these two guys which is in fact the angle that's given to you, the reason why I say is in fact is because one of the trickiest things about work problems is the problem will give you the wrong angle and a lot of people just plug into the equation because they're not going to be careful and it's designed to trick you to do that, right? So, it's going to be this, this is 1000 x the cosine of 37, this is approximately 800 joules that's it.

Alright for part B it says decompose F into Fx and Fy and then find a work done by Fx and Fy, so let me decompose this over here, this is 50 this is Fx, this is 37 and this is Fy right here so Fx is just F cosine of theta, F is 50 cosine of 37 and that is 40. Fy is 50 sine of 37 which is 30, 30 Newtons 40 Newtons so this is 50, 40, 30, and I had to decompose because the problem told me to decompose and then what is the work done by Fx? Well the work done by Fx whatever force this is here just shows up over here, so the work done by a Fx is Fx distance cosine of theta, OK? Now every force has its own angle so you have to refigure out what's the angle but the distance is the same, Ok? Fx is 40 distance is 20, the angle isn't necessarily a 37 because it's a different force, OK? Now let's check this out Fx is moving this way and delta X is moving this way, I hope you agree that the angle between them is 0 so I put a 0 in here different angle and guess what the answer...Ops that's cosine of 0 not just 0 degrees so that's the cosine of 0 and the answer here is 800 as well and the fact that these two answers are the same is not a coincidence they are supposed to be the same but let me keep going here real quick, so for part C I want to know what is the work done by Fy? Fy is 30, the distance is 20, what is the angle? Fy is going up while my displacement is going to the right, the angle between these two is 90 and so I'm going to have here the cosine of 90 and if you plug this into the calculation you should do if you have one on you right now just to check, this is 1 but this is 0 which means there is no work done by Fy, the work done by Fy is 0 and by the way this is because Fy is perpendicular makes a 90 degree angle with the displacement, OK? One last point here that I want to make is that the work done by F if F is a force in 2 dimensions will always be the work done by Fx+ the work done by Fy, most of the time either X....Most of the time one of these two guys will be zero in this case this was 0 and I was left with work of F=work of Fx which is true 800 and 800, OK? So, there's two ways you can do this, you can decompose the vector and then work it out that way when I decomposed Fx the angle between them was now 0 or you can just leave the vector undecomposed and just use the angle between the vector and the displacement, OK? I have a sort of similar practice problem here for you guys to try out, this boat is moving with V up and this person is moving with a V up this way as well so the person's going to move this way and the boat's going to move this way and I want to know how much work is done by the rope in an interval of 5 seconds so it's a little tricky but I think you can figure this out so let's give this a shot.

Practice: A water skier is pulled by a boat (shown). Both move north with a constant 10 m/s. The tension on the rope is 150 N, and makes an angle of 53° with the horizontal. How much work is done by the rope in 5 seconds?

Concept #2: Zero Work, Negative Work & Work by Friction

**Transcript**

Hey guys, so now that we have reviewed the basics of work I want to look into a few special cases of work cases where work is going to be negative or 0 Let's check it out.

So first of all, work and energy are both scalars, right? Remember work and energy are generally the same thing basically the same thing they're both measured in Joules and work is just energy being transferred, they're both scalars which means they have no direction so if you have a 100 joules of work that's not pointing in any direction you just have it, OK? But even though they have no direction they could be positive or negative, OK? Depending on the direction of force, force has direction and displacement, displacement has direction so depending of the direction of those 2-work done by a force can be either positive, negative or 0, OK? So and I think it's helpful if you understand this conceptually so positive work is one when force is done or the force acts in the direction of motion, In direction of motion, negative work is when you're opposite to motion so let's put that in here opposite to motion and if you are neither directly in the direction or nor directly against work if instead you are just sort of to the side if you are perpendicular to the direction of motion, perpendicular has this symbol here 90 degrees then your work will be 0, the force will do no work, OK? One way that you can think about this that I think is useful is positive work is when a force does positive work if it helps motion If a force does negative work if it hurts motion and a force does 0 work if it does not affect the motion at all, OK? So this box here is moving to the right with that's the direction of delta X, let's say it's moving to the right because I'm pushing with a force F and let's say there is also a kinetic friction holding me back a little bit, MG down and normal up, So F is in the direction of motion so I'm going to say that the work done by F would Be positive, friction is opposite to motion so the work done by friction Kinetic is negative normal is perpendicular to the motion I'm moving to the right normal is up that's ninety degrees right there so the work done by normal would be zero and MG is also perpendicular to the motion right so the work done by MG also zero, so positive towards it away from it or against it is negative perpendicular is zero and that it, OK? Let's do an example here and then I'll mention a few more things and then you guys will do practice problem, OK?

A 6-kilogram block sits on a level surface the coefficient of friction between you and surface is 0.7 so let me write this over here, mu is 0.7 between the box and surface, you pull horizontally on it with a force of 50 for a distance of 8 meters, OK? I want to know what is the work done by you? This means the work done by F, OK? Work done by you means the work done by F, well let's see this is F. D cosine of theta, your force is 50 the distance is 8 what about the direction? Well let's try this, I'm pulling this way and it moves this way this angle between these two since they're perpendicular.... IÕm sorry since they're parallel to each other, right? They're parallel to each other like this is 0 so this is going to be the cosine of 0 and here you get 400 Joules, OK? What about the work done by friction? The work done by friction this is kinetic friction because it's going to be sliding there is some friction here, friction kinetic so the work done by the force is that force D cosine of theta so the work done by friction is friction D cosine of theta, but I have to calculate friction so let's do that real quick, friction is Mu Normal I know mu is 0.7 and what about normal? Well normal here is going to be simply MG so I can replace this with MG and then eventually plug in MG so M=6, G we're going to use 10 just for the sake of making this faster and this is 42 Newtons so friction is 42 Notice how I don't plug in friction as a negative remember these forces always get plugged in...F and D always get plugged in to the equation as a positive even if the force is going against you, the distance is 8 and the cosine of theta, now here we have to be careful because we are moving to the right but friction is going to the left and the angle between these two when you have two things like this, OK? In opposite directions the angle between those two is 180 and if you do the cosine of 180 in a calculator it's not 1 it's -1 which means that this is negative whatever these two numbers are so negative -336 Joules, cool?

Part C I want the work done by weight, now the work done by the weight force can be written as the work done by sometimes this is called the work done by gravity but it's really the work done by the force of gravity MG but these are sort of used interchangeably and again the work and by any force is the force MG not just G distance cosine of theta, MG is the mass of 6 so MG is 60 distance is 8 cosine of theta, well I am moving to the right that's my delta X but MG is going down, the angle between these two guys is 90 so this is the cosine of 90 and you put in a calculator and the cosine of 90 is 0 so guess what it doesn't matter what this is the work done by gravity is 0 and the idea here is that gravity doesn't make you faster or slower it doesn't hurt the motion or help the motion it doesn't really effectively do anything to the motion, OK? You could have done this whole thing really much faster by just realizing hey it's perpendicular it makes 90 degree so there is no angle.... IÕm sorry so there's no work that's what's going to happen with normal, normal the work done by normal we're going to know right away it's 0 because normal is perpendicular to my displacement you put a little perpendicular symbol here, OK? And that's it so all of this stuff about positive and negative and zero all come from the fact that you get different angles, here let me write is these angles here real quick, F has an angle of 0 degrees and the cosine of 0 is 1, this guy here has an angle of 90 degrees and the cosine of 90 is 0 so there is no work, this guy makes an angle of 180 degrees with delta X and the cosine of 180 is -1 that's why you get a negative work because the cosine turns out to be negative and here if you go counting from this side here or the shortest the smallest angle between them the angle here again is 90 degrees that's why the work will be 0, cool? So hopefully you got that I want to make a few more points here which is basically just a redo of a recap of some of the stuff that we talked about, so here we did the work done by kinetic friction I want to highlight this here the work done by kinetic friction was FD, OK? Work done by kinetic friction was Fk D cosine of theta but kinetic friction will always oppose motion if you're moving this way kinetic friction is this way and, vice versa right? So, no matter which way you're going if you're going that way kinetic friction is this way so this angle here will always be a 180 for kinetic friction so we don't have to do this over and over again the cosine of 180 is -1 so we can simplify this and say that the work done by kinetic friction is simply -FD now technically that's kinetic friction static friction is different, OK? So, this might be a useful sort of shortcut one really good thing about the equation is that you don't get caught up trying to pick an angle when you're right in the work equation which might end up causing you to pick the wrong angle, right? And then here it's always going to be negative you don't have to worry about angles and whatnot cool. One thing that you should know is the kinetic friction dissipates that's the magic word dissipates mechanical energy into heat, OK? I mentioned earlier but just revisiting here heat and heat is thermal energy so it dissipates it burns mechanical energy into heat, OK? Some questions will ask you how much mechanical energy was lost by an object? Or how much energy was dissipated (dissipated again comes from here) by an object? And the idea is that in this problem up here F gave me 400 Joules of energy and the work done by friction was -336 which means friction stole took away 336 joules of energy from the system, so that's the amount of energy dissipated the amount of energy dissipated is the work done by kinetic friction so you just have to sort of translate that your head now that being said if my work was -336 I don't say that I lost -336, right? If you lose 10 bucks you wouldn't say I lost -10 dollars you just say I lost 10 dollars so this is going to be the positive work, OK? So, in this case how much energy did you lose you wouldn't say I lost -336 you'd say I lost 336 so it's the positive of the work done by friction which is this equation right here which you can also think of it as just simply FD the work done by friction is -FD, the amount of energy dissipated +FD, OK? The last point that I want to make here before you go into this practice problem is some forces will never do work in other words the work done by those forces is always going to be 0, why? Because they're always perpendicular to the motion normal by definition that's what normal means it's perpendicular to the surface and if you're moving along the surface normal is up perpendicular right no matter where you are normal is always perpendicular, centripetal forces again by definition centripetal forces if you go in a circle you're going this is the direction the instantaneous direction of your displacement you're temporarily going that way and obviously you readjust but then some force will pull you to the middle here whatever force that is any force that acts as a centripetal force does no work so if I'm spinning something with a tension the work done by tension is 0 that's because they make 90 degrees everywhere you are, same thing with MGY, MGY is always opposite to normal right here so the work done by MGY is 0. So, I think it's helpful if you remember these because you can knock them out faster, OK? So, from now when we get asked these questions I'm just going to say 0 because you're expected to remember that, OK? And I might from time to time remind you that it's because it's perpendicular where it makes 90 degrees with the displacement, cool? Let's try this practice problem here I have a box on the surface and then we're going to pull the box it starts moving whatnot and I want to know the work done by all these different forces here and then there's a bonus question here let's give this a shot.

Practice: A 3-kg box sits on a level surface. The box-surface coefficient of friction is 0.4. You pull on the box with 20 N at 37o above the horizontal for 5 meters. Calculate the work done by:

(a) you;

(b) friction;

(c) weight; and

(d) normal.

How much energy was dissipated? Where did this energy “go” ?

0 of 6 completed

A 50 kg crate is at rest on the floor, and you need to move it across a room. The floor you are moving it across has a coefficient of static friction of 0.5 and a coefficient of kinetic friction of 0.3. In order to move it across the room, first you need to move it 5 m to the left, then 7 m forward, then finally 3 m to the right. How much work do you do on the box in order to move it across the room?

An object moves along a floor with some coefficient of kinetic friction. Which of the following statements is true?
a. The work done by friction is positive, and doesn’t depend upon the path taken
b. The work done by friction is positive, and does depend upon the path taken
c. The work done by friction is negative, and doesn’t depend upon the path taken
d. The work done by friction is negative, and does depend upon the path taken

A fisherman reels in 12.0 m of line while pulling in a fish that exerts a constant resisting force of 25.0 N.If the fish is pulled in at constant velocity, how much work is done on it by the tension in the line?

You and three friends stand at the corners of a square whose sides are 8.0 m long in the middle of the gym floor, as shown in the figure . You take your physics book and push it from one person to the other. The book has a mass of 1.6 kg , and the coefficient of kinetic friction between the book and the floor is k=
0.26.The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement?You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement?You slide the book to Kim who then slides it back to you. What is the total work done by friction during this motion of the book?Is the friction force on the book conservative or nonconservative?Explain.

A 30.0-kg packing crate in a warehouse is pushed to the loading dock by a worker who applies a horizontal force. The coefficient of kinetic friction between the crate and the floor is 0.20. The loading dock is 15.0 m southwest of the initial position of the crate.If the crate is pushed 10.6 m south and then 10.6 m west, what is the total work done on the crate by friction?If the crate is pushed along a straight-line path to the dock, so that it travels 15.0 m southwest, what is the work done on the crate by friction?

A 0.60-kg
book slides on a horizontal table. The kinetic
friction force on the book has magnitude
1.6 N .How much work
is done on the book by friction during a displacement of
2.7 m
to
the left?The book now slides
2.7 m
to the right, returning to its
starting point. During this second
2.7-m
displacement, how much
work is done on the book by friction?What is the total work
done on the book by friction during the complete round trip?On the basis of your answer to part C, would you say that the
friction force is conservative or nonconservative?

A large crate sits on the floor of a warehouse. Paul and
Bob apply constant horizontal forces to the crate. The force applied by Paul has magnitude
51.0 N
and direction 61.0
south of west.How much work does Pauls force do during a displacement of the crate that is 12.0 m
in the direction 22.0
east of north.

A child is pulling a wagon down the sidewalk. For 9.0 m the wagon stays on the sidewalk and the child pulls with a horizontal force of 23 N . Then one wheel of the wagon goes off on the grass so the child has to pull with a force of 38 N at an angle of 13 to the side for the next 5.5 m . Finally the wagon gets back on the sidewalk so the child makes the rest of the trip, 13.0 m , with a force of 23 N .How much total work did the child do on the wagon?

A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of 15.015.0 with her direction of motion, and then continues in a straight line. The tension in the rope is 180 N.</noba>How much work is done on the skier by the rope during a displacement of 300 m?

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106 N , one an angle 15 west of north and the other an angle 15 east of north, as they pull the tanker a distance
0.76 km
toward the north.What is the total work they do on the supertanker?

Two men, Joel and Jerry, push against a concrete wall that is 3 meters thick. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. How does the work that Joel does on the wall compare to the work that Jerry does on the wall?A) Both men do positive work, but Jerry does 50% more work than Joel.B) Both men do positive work, but Joel does 25% more work than Jerry.C) Both men do positive work, but Joel does 50% more work than Jerry.D) Both men do positive work, but Joel does 75% more work than Jerry.E) Neither of them does any work.

The three ropes shown in the birds-eye view of the figure are used to drag a crate 3.6 m across the floor. How much work is done by each of the three forces?

Using a cable with a tension of 1310 N, a tow truck pulls a car 5.00 km along a horizontal roadway. How much work does:(a) the cable do on the car if it pulls horizontally(b) the cable do on the car if it pulls at 35.0° above the horizontal(c) gravity do on the car in part A(d) the cable do on the tow truck if it pulls horizontally(e) the cable do on the tow truck if it pulls at 35.0° above the horizontal

A 2400 N crate rests on the floor. How much work is required to move it at constant speed:(a) 4.1 m along the floor against a drag force of 240 N(b) 4.1 m vertically

An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.30 m deep, but survived with only minor injuries. Assuming the pilots mass was 88.0 kg and his terminal velocity was 46.0 m/s, estimate the:(a) work done by the snow in bringing him to rest (b) force exerted on him by the snow to stop him(c) work done on him by air resistance as he fell

A shopper in a supermarket pushes a cart with a force of 35.0 N directed at an angle of 25.0° below the horizontal. The force is just sufficient to balance various friction forces, so the cart moves at constant speed. (a) Find the work done by the shopper on the cart as she moves down a 50.0-m-long aisle. (b) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the friction force doesn’t change, would the shopper’s applied force be larger, smaller, or the same?(c) What about the work done on the cart by theshopper?

A raindrop of mass 3.35 x 10-5 kg falls vertically at constant speed under the influence of gravity and air resistance. Model the drop as a particle. As it falls 100 m, what is the work done on the raindrop (a) by the gravitational force and (b) by air resistance?

On a farm, you are pushing on a stubborn pig with a constant horizontal force with magnitude 30.7 N and direction 37.0 counterclockwise from the +x+x-axis.(a) How much work does this force do during a displacement of the pig that is s⃗ = (5.00 m) î?(b) How much work does this force do during a displacement of the pig that is s⃗ = –(6.00 m) ĵ?(c) How much work does this force do during a displacement of the pig that is s⃗ = –(2.00 m) î +(4.00 m) ĵ ?

A physics student uses a force P of magnitude 70N and angle = 50 degrees (with respect to the horizontal) to push a 5.2 kg block across the ceiling of her room, as shown in the figure to the right. The coefficient of kinetic friction between the block and the ceiling is 0.35.A. Draw a free body diagram of the systemB. What is the magnitude of the block's acceleration?

Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.6 m/s how much work does Sam do to stop it?

A 710 kg car drives at a constant speed of 23 m/s. It is subject to a drag force of 500 N. What power is required from the car's engine to drive the car (a) on level ground? (b) up a hill with a slope of 2.0°?

Find the work W done by the 18-newton force.

What is the transfer of energy to a system by the application of a force called?a) Dot product b) Power c) Work d) Watt e) Energy transformations

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.90×106 N, one at an angle 19.0 west of north, and the other at an angle 19.0 east of north, as they pull the tanker a distance 0.870 km toward the north.What is the total work done by the two tugboats on the supertanker?Express your answer in joules, to three significant figures.

The two ropes seen In the figure are used to lower a 255 kg piano 6.00 m from a second-story window to the ground.How much work is done by each of the three forces ?

Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacementIf an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated asW= F*s=/F/ /s/ cosθwhere W is the work done by force F on the object that undergoes displacement s directed at angle θ relative to F.Note that depending on the value of cosθ, the work done can be positive, negative, or zero.In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. (Intro 1 figure)Find the work W done by the 30-newton force.Use two significant figures in your answer. Express your answer in joules.

Find the work W done by the 12-newton force and the 15-newton force

How much work is done by the boy pulling his sister d = 27.0 m in a wagon as shown in figure below? Assume no friction acts on the wagon and he pulls with a force of 62.0 N and at an angle θ = 31.0°.

How much work is done by the boy pulling his sister d = 30.0 m in a wagon as shown in figure below? Assume no friction acts on the wagon and he pulls with a force of 50.0 N and at an angle θ = 30°.

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