0
$\begingroup$

I don't really understand the difference and the relation between work and kinetic energy. When you move an object a distance you do work (or does the object work?), what's the object's kinetic energy? Is the kinetic energy the work you've put in to the box all gathered up at the end of the distance given the box keeps moving when you stopped pushing the box? The box can't have kinetic energy if's stationary after I've pushed the box can it?

Sorry if it's unclear.

$\endgroup$
  • $\begingroup$ Have you read up the definitions? $\endgroup$ – gented Jun 6 '17 at 15:29
  • $\begingroup$ Yeah, just didn't really understand them @GennaroTedesco $\endgroup$ – Iram Haque Jun 6 '17 at 15:34
  • $\begingroup$ What's there to understand? The kinetic energy is defined as $1/2 mv^2$ and the work done by a force along a path is equal to the difference in kinetic enery between the endpoints. $\endgroup$ – gented Jun 6 '17 at 15:35
  • $\begingroup$ Well, I didn't quite pick that up. @GennaroTedesco $\endgroup$ – Iram Haque Jun 6 '17 at 15:39
0
$\begingroup$

The kinetic energy is indeed the work you have done (assuming no change in potential energy), however this is also neglecting the effect of friction. In real life, the kinetic energy of the box is going to be constant if its moving at a constant velocity even if you are exerting a force on the box and hence doing work on the box. This is because in this case, the work you are doing is being converted into heat and sound due to friction. If we neglect friction, your box would keep accelerating as you applied a force and hence the kinetic energy would keep increasing.

$\endgroup$
  • $\begingroup$ So not considering friction, if you've don a certain amount of work on a object, that amount of work is equal to the kinetic energy, correct? $\endgroup$ – Iram Haque Jun 6 '17 at 15:27
  • $\begingroup$ Yes, although this also assumes there is no change in the potential energy of the object. $\endgroup$ – EigenFunction Jun 6 '17 at 15:39
0
$\begingroup$

Work translates directly into kinetic energy unless there is a potential field or friction. In the case of a potential the energy of work may be shared between kinetic energy and potential energy. When friction is present some or all of the energy of work may go into heating the environment.

$\endgroup$
  • $\begingroup$ So it will reduce the kinetic energy but surely the work-energy theorem will apply,right? $\endgroup$ – Hydrous Caperilla Jan 13 '18 at 2:34
  • $\begingroup$ On second thought it won't ....got that' $\endgroup$ – Hydrous Caperilla Jan 13 '18 at 2:36
-1
$\begingroup$

When we move an object, we apply a force. We do not do any work nor does the box do any work. The force does work. The difference between work and energy is that energy is the capacity to do work. You can think of energy as stored work. The relation between work and energy is that the work a force does on an object equals the change in energy, not just kinetic energy but any form of energy. The box cannot have kinetic energy if it is stationary but it acquires kinetic energy when a force does work on it and it continues to be in motion.

$\endgroup$
  • $\begingroup$ what happens to energy when it's stationary, it doesn't disappear right? $\endgroup$ – Iram Haque Jun 6 '17 at 15:29
  • $\begingroup$ You have to do work to stop an object in motion. The sign(positive or negative) of work done to stop an object is opposite to that of the work done to put it in motion while the magnitude remains the same. So, the total work becomes 0. $\endgroup$ – Harmohit Singh Jun 6 '17 at 15:39
  • $\begingroup$ If we'd consider a particle with no friction, if a force has earlier affected it and right now the particle is stationary, does it still have energy? $\endgroup$ – Iram Haque Jun 6 '17 at 17:58

protected by Qmechanic Jun 6 '17 at 15:35

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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