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Thought experiment:
A block with a mass of 10kg is moving right with velocity 1m/s. I pushed the object leftwards so that its velocity changed to -1m/s.

In this case, the kinetic energy of the block remains constant(5J) before and after the push. Does this mean that no work is done on the block? This feels very strange, as I applied force on the object and caused displacement, just that its direction is opposite. Is there something wrong with this approach?

Here are some points worth considering:

  • Kinetic energy is scalar, so the direction the object is moving in doesn’t matter. (Kinetic energy can’t be negative)
  • Velocity is vector, so of course my hands did apply a leftward force to the block, which resulted in negative acceleration.
Jongwoo Lee
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    It simply means that you are picking a special corner case without considering that the block had to have a velocity of 0m/s in that rest system during some part of that process. If you define energy as the ability of the system to perform work on another system, then the total energy first decreases, then increases again. This is with the caveat that kinetic energy is observer dependent, of course. A better way of looking at this is that energy is always a system property rather than a property of one object alone. – FlatterMann Jun 09 '23 at 10:38
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    An object moving in a circle at constant speed changes its velocity continually even though no work is done on it. – gandalf61 Jun 09 '23 at 10:44
  • @gandalf61 Oh sorry for the misleading title, I was only considering cases where F and S are not perpendicular. – Jongwoo Lee Jun 09 '23 at 11:22
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    “This feels very strange, as I applied force on the object and caused displacement” During exactly half of the process, the object imparted displacement to you. That provides the symmetry. – Chemomechanics Jun 09 '23 at 13:57
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    Imagine that "you" actually are a spring. The kinetic energy of the moving object would be "absorbed" by the spring and stored in the spring as the object's speed slows to zero, and then the spring returns that energy when the object begins to move in the opposite direction. But wait! You are not a spring? You can't "store" and "return" the energy? Well that just means that you are not as efficient as the spring. You allowed the energy to be turned to heat. – Solomon Slow Jun 09 '23 at 15:23
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    Another physical example of this is the Newton's cradle. A ball approaches the central group of balls at x m/s, its motion is arrested , sending the ball on the other side into motion. Then that ball comes back, hits the central group again, which knocks the original ball bacwards at exactly -x m/s. No energy entered or left the system (well, small friction losses, of course) – Cort Ammon Jun 09 '23 at 22:20

2 Answers2

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This is a funny problem. Thanks for posting it. So here is the explanation. You have the work kinetic energy theorem that states the following:

$W = \Delta K$

where $K$ is the kinetic energy. So, as you mentioned it is true that the variation of the kinetic energy is zero. The theorem then states that the work of the force you used to make the change is also zero. So, you are right. You applied a force and it made no work. But let us examine in a little detail why is that.

We may assume a constant force $F$ is applied. So, you apply this force contrary to the direction of motion (to the left, assuming object is moving initially to the right). The velocity starts to decrease to zero. Let us assume that in that process the object moved a distance $d$ to the right. The work done in that part is

$W_{\text{right}} = - F d$ (force is in opposite direction to displacement)

After that, with the force applied in the same direction the object accelerates to the left. And needs a distance $d$ to get to the velocity $- 1 \text{m/s}$. It then does a work of:

$W_{\text{left}} = F d$ (force is in the direction of displacement)

So, the total work is the sum of the two contributions:

$W = W_{\text{left}} + W_{\text{right}} = F d - F d = 0$

Hope that helps!

Amit
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Blue
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    Thanks, that explains it. Would this be also true for collisions, e.g. if it crashed into a wall and changed directions? – Jongwoo Lee Jun 09 '23 at 11:06
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    @JongwooLee Yes, if the KE doesn't change, the net work must be zero. You can passively bounce a ball off a wall, no energy input is required to undergo a collision (imagine if bouncy balls needed to be battery powered!). As another example, consider swinging an object on a string in a circle over your head - the string isn't doing any work, but the velocity points in opposite directions half a rotation apart. – Nuclear Hoagie Jun 09 '23 at 16:18
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    The net work, i.e. the total work done, must be zero: this is the key. You do do some work: some of it is positive and some of it is negative. It is just that in your example the positive and the negative parts cancel exactly. – printf Jun 09 '23 at 19:14
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Work is the integral over time.

Where the equation for energy use is linear, you can replace this with the difference between start and end. With higher-order equations though, non-linear behaviour during the move means you can't make that assumption.

Graham
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