0
$\begingroup$

If work done equals change in kinetic energy and change in gravitational potential energy, is any work done if we move a standing object and at end point its speed is 0 km/h and height remains same?

On the other hand, work equals force times displacement. We did change its place and applied force, which implies that the work was done.

These contradict in my current understanding.

Thanks for answers!

$\endgroup$
2
  • $\begingroup$ You need to be more specific here. You are just saying "work", but you need to specify which force is doing the work. $\endgroup$ Commented Jun 2, 2021 at 0:48
  • $\begingroup$ I agree with @BioPhysicist. In addition, is friction involved when you change an object's place with an applied force? $\endgroup$ Commented Jun 2, 2021 at 1:01

3 Answers 3

5
$\begingroup$

There was a work $w$ to accelerate the object to a given velocity, and another work $-w$ to decelerate it to zero. The net work is zero.

Remember that $dw = \mathbf {F.dx}$. When F changes direction the dot product changes sign.

$\endgroup$
3
$\begingroup$

First, there are two key statements that are important to keep separate.

  1. The work-energy theorem says that the net work is equal to the change in kinetic energy.

  2. The work done by a gravitational field, is equal to the minus the change in the gravitational-potential energy.

The way I interpret your question, is that you move a block on a horizontal table from point $A$ to point $B$, in such a way that its initial and final velocity are $0$, and that its height doesn't change.

Then the gravitational field does no work, $W_{g}=0$ (since the height of the block never changes). Furthermore the difference in gravitational potential energy between points $A$ and $B$ is zero, $U_g(B)-U_g(A)=0$. Statement 2 reads $W_g=-(U_g(B)-U_g(A))$. Since $0=0$, there is no contradiction.

I think you may also be confused because there is non-zero work done by whatever is pushing the block as it moves from $A$ to $B$. This is true, but there is a zero net work. In other words: while moving from $A$ to $B$, obviously the block has some non-zero velocity and so it has some kinetic energy. Some work is needed to speed up the block from $0$ velocity at point $A$, to a non-zero velocity. But then some other work is needed to stop the block as it approaches point $B$. Since the change in kinetic energy is $0$ (the block started and stopped at rest), the net work must be zero. In other words, the sum of the work done by the force that speed the object up, and the work done by the force that stopped the object, must exactly cancel. This is not an obvious statement, given that completely different forces could be responsible for speeding up the block and slowing it down (maybe a rocket gave the block and initial boost and friction ground it to a halt). But this is the power of the theorem. You can check in special cases that the work energy theorem does in fact work -- for example, if you attach a block to a spring, pull on the block, and release it from rest, you can check explicitly that the work the spring does speeding up the block exactly cancels with the work the spring does in slowing it down as it approaches its resting position.

$\endgroup$
0
2
$\begingroup$

When you move the object, it must have some speed. So in this case you have to accelerate the object and decelerate to get 0 speed at the endpoint. Thus when you calculate work done by a change in kinetic energy it's zero showing that you apply work W to accelerate and -W to decelerate the object.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.