Work done under constant velocity

[A.T.]

Remember that scalar quantities are directionless, so summing them up as if they were vectors is an incorrect concept.

Then try to convince your bank that they should stop counting the withdrawals as negative for your account's balance.

 

[OldYat47]

OK, I've spent some time stripping this to the simplest possible analysis, and this is my last post, promise.
"Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work."
Just like your spring.

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction. The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0. Suppose the observation stops when the block is at its original location. Positive work is done at both ends since the force at both ends is in the same direction as the displacement of the block.

So the sum of the work is (2 X work), not zero.

 

[jbriggs444]

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction. The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0.

That is not the scenario under consideration. The situation under consideration (as described in the original post) contemplates two equal and opposite forces applied at the same time while the mass is moving. One is applied in the direction of motion. Since the other is opposite, it is clearly applied opposite to the direction of motion.

It is a mathematical fact that the work done by equal and opposite forces on the same object at the same time will be equal and opposite.

[For point-like objects anyway. There are caveats for non-rigid or rotating extended objects]

 

[Doc Al]

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction.

OK.

The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0.

Not going to happen. Realize that even though the second push is in the -x direction, the block is still moving in the +x direction until it is brought to a stop.

Suppose the observation stops when the block is at its original location. Positive work is done at both ends since the force at both ends is in the same direction as the displacement of the block.

You need to rethink this.

 

[Dale]

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction. The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0.

First, this is impossible. If the v is in the +x direction and a force is applied in the -x direction to bring it to -v then the box must be brought to v=0 first.

It is possible to apply a force to bring it from v to -v without going to v=0, but in that case the force cannot be strictly in the -x direction. Instead, the force must turn the object, typically by always acting perpendicular to the velocity.

Positive work is done at both ends since the force at both ends is in the same direction as the displacement of the block.

This is not true in general.

Take the case of the turn. In that case the force is always perpendicular to the velocity so the dot product is 0 and no work is done at all.

Now, take the case of the straight line force. From when the velocity is v until it is 0 the force is in the opposite direction of the velocity, so the dot product is negative and negative work is done. Then from when the velocity is 0 until it is -v the force is in the same direction as the velocity, so the dot product is positive and positive work is done.

In all cases, the net work at the second end is 0. (Assuming no friction etc)

So the sum of the work is (2 X work), not zero.

No, in your notation the sum of the work is (1 x work).