0
$\begingroup$

I am a bit troubled trying to make connection between energy and work. Image there is a block on frictionless road on x axis with initial velocity of 0 and a force F is applied for some distance d. the work done is equal to change in kinetic energy and this block has some kinetic energy now and with velocity V. What I do not understand is that this block has energy which means it has ability to do work but how does it do that? Where is the work done by the block? It is just going on x axis. I am confused a lot.

$\endgroup$
3
  • $\begingroup$ @StudyStudy do you mean once the block is loving, it has /kinetic/ energy? $\endgroup$ – electronpusher Jan 30 '20 at 21:26
  • $\begingroup$ Feynman is very good on energy and work, I think: it's at least worth reading the start of that lecture, and probably all of it. $\endgroup$ – user107153 Jan 30 '20 at 21:30
  • $\begingroup$ @electronpusher, Thanks, I worded my comment in a confusing way so I deleted it. But I feel that OnurTR must be clear on the difference between kinetic and potential energy and on the distinction between work done "on" an object and work done "by" the object. $\endgroup$ – StudyStudy Jan 30 '20 at 21:43
2
$\begingroup$

I am a bit troubled trying to make connection between energy and work.

Work is one of two means by which energy can be transferred from one object to another by means of a net force acting on an object through a distance. The other means of energy transfer is heat, which is energy transfer due solely to a temperature difference between objects.

What I do not understand is that this block has energy which means it has ability to do work but how does it do that?

When the force $F$ caused the block to accelerate from 0 to some velocity $v$ the force has transferred energy from the source of the force to the block giving the block kinetic energy. Now that the block has kinetic energy with respect to the frame of reference where it had none, it (the block) now has the ability to transfer energy to something else.

Now suppose that block, which now has kinetic energy of $\frac{mv^2}{2}$ collides with another block at rest with the same mass in a perfectly elastic collision. The first block stops and the second block continues on with the same velocity and kinetic energy as the first block had before the collision. The first block has transferred its kinetic energy to the second. It has done work on the second block.

The impact force of block 1 on block 2 transfers the energy. Per the work energy theorem the work done by block 1 on block 2 is

$$W=F_{ave}d=\frac{mv^2}{2}$$

Where $F_{ave}$ is the average impact force block 1 exerts on block 2, and $𝑑$ is the stopping distance of block 1. What the individual values actually are would be would depend on the elastic properties of the materials involved (e.g, modulus of elasticity). The “springier” the materials the lower $F_{ave}$ and the larger $d$. You can also say that block 2 does an equal amount of negative work on block 1, taking energy from block 1.

Hope this helps.

$\endgroup$
5
  • $\begingroup$ yes thanks a lot $\endgroup$ – OnurTR Jan 30 '20 at 22:03
  • $\begingroup$ Excellent answer, +1. So if the force does not transfer any energy then how much work does it do? $\endgroup$ – Dale Jan 30 '20 at 22:32
  • $\begingroup$ @Dale The impact force of block 1 on block 2 transfers the energy. Per the work energy theorem the work done by block 1 on block 2 is $W=F_{ave}d=\frac{mv^2}{2}$ where $F_{ave}$ is the average impact force block 1 exerts on block 2, and $d$ is the stopping distance of block 1. What individual values actually would be would depend on the elastic properties (e.g, modulus of elasticity) of the materials involved. The “springier” the materials the lower $F_{ave}$ and larger $d$. $\endgroup$ – Bob D Jan 30 '20 at 23:02
  • $\begingroup$ @OnurTR I've added some more details concerning the work involved. Hope it helps. $\endgroup$ – Bob D Jan 30 '20 at 23:21
  • $\begingroup$ @Bob D thanks. But just hypothetically IF some force (not necessarily an impact force) doesn’t transfer any energy then how much work does it do? $\endgroup$ – Dale Jan 30 '20 at 23:42
1
$\begingroup$

"What I do not understand is that this block has energy which means it has ability to do work but how does it do that?" Lasso the block with a rope and hang on to the rope. The block will come to rest exerting a force on you, the holder of the rope, through a distance as your hand is pulled forward. Work is done on your hand.

There are, of course many other ways in which the moving body can be made to do work, but the essential feature is that a force from an external object on the block is needed to bring it to rest, and using Newton's third law, the block exerts a force on that object – and moves it through a distance.

What's more we can show that if the block (of mass $m$) is moving at speed $v$, the amount of work it does in coming to rest is $\tfrac12 mv^2$.

$\endgroup$
2
  • $\begingroup$ However, in collisions of two objects(one moving, one stationary) generally exertion time of force applied from moving object to another lasts so small and we do not really see a distance the second stationary object moves during collision. It just happens so fast and that is why force and work seem vague a bit. I know energy is always preserved during collisions. By the way I want to ask that (rope example) is the work done on me equal to work that I have done on the object if the object comes to rest because I also apply force on the object through the rope? $\endgroup$ – OnurTR Jan 30 '20 at 21:47
  • $\begingroup$ @OnurTR the person stopping the block does negative work on the block because the force he/she exerts is opposite the displacement of the block while coming to a stop ( the change in kinetic energy is negative). Negative work means the person takes energy away from the block instead of giving it energy. So the energy is transferred from the block to the person. $\endgroup$ – Bob D Jan 30 '20 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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