High school physics
Work example problems
David goes through some example problems on the concept of work. Created by David SantoPietro.
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- At4:33he says no net work due to no change in kinetic energy. Is there not a change in potential energy that would account for work being done?(70 votes)
- If the trashcan's velocity is constant, that means it has no net acceleration. If it has no acceleration, there is no net force on the object, since F=ma=4kg(0)=0. If there is no net force, then no net work is being done on the object, since W=fd=0(2m). However, you are accelerating the trashcan, applying a force to it, and doing work on it when you lift it, it is just that all of your work is being negated by gravity in this example of constant velocity. Hope this helps!(7 votes)
- At4:06, I don't understand why the force lifting the trash can up is 39.2N. If it's 39.2N up and 39.2N down, would the forces balance on both sides and the trash can would not be lifted up? Doesn't the upward force have to be greater in some way for it to move in that direction?(51 votes)
- Yes, you need slightly more than 39.2N to trigger the initial acceleration.(5 votes)
- He uses kinetic friction in the video but how would it be effected if he also had Static Friction?(11 votes)
- static friction applies when there is no movement. Hence, no displacement = no work done.(36 votes)
- But wait. I know that the net force needs to be zero for avoiding any acceleration, but if there was no budging force in the beginning, why would the trashcan even go in the air with a constant velocity? Both forces are equal, so it'd rather stay on the ground.
I do understand that Fn = 0 makes sense when it's already in a velocity, but how does it make sense when it was at rest and went into a constant motion just because a force equal to gravity began pushing it from below?(11 votes)
- Well we assume the body is already in motion.So we ignore the fact that an initial force was applied which was larger than gravitational force and whose acceleration was used to attain the constant velocity.(8 votes)
- 5:31will a very less force just greater than zero can lift the block because normal from the ground will be balancing mg and it could move up with a little force(4 votes)
- Good question. The answer is no. It might seem that the laws of physics say yes, because if normal is equal and opposite to mg, then adding a small upward force will result in a net upward force. So what happens? Well, the normal force is a funny kind of force, which you might call a "reactive" force -- it only pushes exactly hard enough to keep the object from pushing through the surface, but no harder. So if a 100N block is sitting on a horizontal table, the normal force will be 100N, up. If you pull upwards with a force of 2N, the normal force instantaneously (well, almost) drops to 98N, so that the net force is still zero, and the block does not move. If you increase your upward pull to 99N, the normal force will drop to 1N. This should agree with common sense and your experience -- if you want to lift a 100N object, YOU have to pull upward with 100N of force! (note: you mention the time5:31in your question, but this video is only4:49in total length!)(16 votes)
- How is there no change in KE in the example, at the very end of the video (4:44). Initially velocity was zero and then after a 2 mt displacement, it had some velocity, so there has to be a change in KE?(4 votes)
- Yeah, this is a bad example. He needed to make one of the following conditions:
1) the can starts from rest and ends at rest. In that case you put in some extra work to accelerate the can but then you get back the same amount when you decelerate it, so the total work is just F*d
2) the can somehow already had upward velocity and all we did was maintain that upward velocity to get it to the new height. That's sort of what he implied when he said the can had constant velocity, but it doesn't make practical sense because the can started on the ground.
A better example might to put the can in an elevator. The elevator starts at the first floor and stops at the 4th floor. How much work was done on it between floors 2 and 3 when it moved at constant velocity? That's f*d.(15 votes)
- At1:50, how can the work done be a negative value when work is a scalar quantity? I understand that cosine of 180 is -1, but is the negative value still relevant?(5 votes)
- Yes. It means that the force and displacement vectors are opposite. So, lifting a crate up means gravity is doing negative work and you do positive work. If you lower a crate, the opposite is true.(8 votes)
- At2:15why is the angle listed at 90 degrees and not 270 degrees? I'm watching all of the Khan Academy videos on physics, and in an earlier physics video on vectors, they said that vector angles are measured in a counter clockwise direction starting from the east.(5 votes)
- It does not matter. There certainly is no law of physics that says angles must be measured that way.(4 votes)
- At4:42, what if the velocity was not constant when it moved 2m up (but even after 2m it was still moving) what would be the work then?(4 votes)
- Well then the work would be calculated like the example given before using KE= 1/2 (mv^2). We would need to know the velocity to calculate the KE (correct me if I'm wrong)(2 votes)
- So wait a minute, how does an object gain kinetic energy if, like in the second example (pushing the trashcan upwards), there is no net work done on the object? How does an object gain kinetic/potential energy at all in these circumstances? #lateatnightphysicsthoughts Seriously, though, I am alarmed at my confusion over this.(4 votes)
I'm going to show you some examples of how to solve problems involving work. Imagine a 4 kilogram trashcan. The trashcan is disgusting. So someone ties a string to it and pulls on the string with a force of 50 newtons. The force of kinetic friction on the trashcan while it slides is 30 newtons. The trash can slides across the ground for a distance of 10 meters. Let's try to find the work done by each force on the trash can as it slides across the ground. To find the work done by each force, we should recall the formula definition of work. Work equals Fd cosine theta, where theta is the angle between the force doing the work and the direction the trashcan is moving. There are four forces involved here-- tension, the normal force, the gravitational force, and the force of kinetic friction. In finding the work done for all of these forces, the displacement is going to be 10 meters. But the value of the force and the angle between that force and the displacement is going to differ for each of the forces. For instance, to find the work done by the force of tension, we'll plug in the size of the tension, which is 50 newtons. The displacement is 10 meters. And since the tension force is pointed in the same direction as the displacement, the angle between the force of tension and the displacement is 0 degrees. And since cosine of 0 is 1, the work done by the tension force is 500 joules. To find the work done by friction, we'll plug in the size of the force of friction, which is 30 newtons. The displacement is still 10 meters. And since the force of friction points in the opposite direction as the displacement, the angle between the force of friction and the displacement is 180 degrees. Since cosine of 180 is negative 1, the work done by the force of friction is negative 300 joules. Now let's figure out the work done by the gravitational force. The force of gravity is mg. So the force of gravity is 4 kilograms times 9.8 meters per second squared, which is 39.2 newtons. The displacement is again 10 meters. But the angle between the gravitational force and the direction of the displacement is 90 degrees in this case. And since cosine of 90 is 0, the gravitational force does no work on this trashcan. Similarly, if we were to find the work done by the normal force, the angle between the direction of the displacement and the normal force is 90 degrees. So the normal force also does no work on the trashcan. This makes sense because forces that are perpendicular to the motion can never do any work on that object. So that's how you can find the work done by individual forces. And if we wanted to know the net work done on this trashcan, we could just add up the work done by each individual force. So the net work is going to be 200 joules. Now that we know the net work done on the trashcan, we can use the work-energy principle to figure out the speed of the trashcan after it's slid the 10 meters. The work-energy principle says that the net work done on an object is equal to the change in kinetic energy of that object. So 200 joules is going to equal the difference in kinetic energy. If we assume the trashcan started at rest, which seems reasonable, the initial velocity is 0. So we can solve for the final speed of the trashcan, which comes out to be 10 meters per second. This time, let's say you take the trashcan and lift it upwards with a constant velocity for a distance of 2 meters. In order to lift the trashcan up with a constant velocity, you need to push with a force equal to the weight of the trashcan, which means you have to push upwards with a force of 39.2 newtons. So to find the work done by the force that you exert, the force is going to be 39.2 newtons. The displacement is going to be 2 meters. And the angle between the force and the displacement is going to be 0 degrees because the direction of the force that you exert is in the same direction as the displacement of the trashcan. So the work that you've done in lifting up this 4 kilogram trashcan is 78.4 joules. To find the work done by the force of gravity, we can use the force of gravity, which is again 39.2 newtons. The displacement is again 2 meters. But the angle between the direction of the displacement and the gravitational force is 180 degrees because the displacement points up and the gravitational force points down. So the work done by the gravitational force is negative 78.4 joules, which means the net work done on the trashcan is 0. And that makes sense. Because since the trashcan moved upwards with constant velocity, there was no change in the kinetic energy of this object.