Question 1

At 4:33 he 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?

Accepted Answer

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!

Question 2

At 4:06, 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?

Accepted Answer

Yes, you need slightly more than 39.2N to trigger the initial acceleration.

Question 3

He uses kinetic friction in the video but how would it be effected if he also had Static Friction?

Accepted Answer

static friction applies when there is no movement. Hence, no displacement = no work done.

Question 4

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?

Accepted Answer

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.

Question 5

At 1:50, 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?

Accepted Answer

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.

Question 6

At 2:15 why 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.

Accepted Answer

It does not matter.  There certainly is no law of physics that says angles must be measured that way.

Question 7

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?

Accepted Answer

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
or
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.

Question 8

5:31 will 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

Accepted Answer

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 time 5:31 in your question, but this video is only 4:49 in total length!)

Question 9

At 4:42, 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?

Accepted Answer

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)

Physics library

Course: Physics library > Unit 5

Work example problems

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