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Let's assume that a ball with some mass is moving horizontally with some constant velocity.
A constant force is acting on the ball in a direction opposite to that of the motion of the ball.
Now, the ball will decelerate and eventually come to rest.

In this scenario, did the ball do any work?

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  • $\begingroup$ Is it important that it is a ball? The shape doesn't seem relevant to your description $\endgroup$
    – Steeven
    Jun 25, 2020 at 10:51
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    $\begingroup$ It isn't. I just took it as an example. $\endgroup$
    – Rajdeep Sindhu
    Jun 25, 2020 at 10:53
  • $\begingroup$ you need to provide more information about that constant force. $\endgroup$
    – Bhavay
    Jun 25, 2020 at 11:04
  • $\begingroup$ @Bhavay I don't understand what you mean, what kind of information shall I provide? $\endgroup$
    – Rajdeep Sindhu
    Jun 25, 2020 at 11:06
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    $\begingroup$ Then you already have your answer sir. $\endgroup$
    – Bhavay
    Jun 25, 2020 at 11:21

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No, the ball did no work.

Also note, it is not objects that do work in a physical sense, it is forces that do work.

If you throw a ball up in the air, then during the flight only gravity acts. The force of gravity (weight) is the only force, so this is the only force that can do work. And it is doing work, by converting kinetic energy into gravitational potential energy.

The ball has no influence on the work done.

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  • $\begingroup$ If only forces are said do work, what does this definition of energy mean : "An object is said to possess energy if it has the ability to do some work". That's the definition in my last year's textbook. This is also the definition that was taken into context while deriving the formula for Kinetic Energy. $\endgroup$
    – Rajdeep Sindhu
    Jun 25, 2020 at 11:00
  • $\begingroup$ @RajdeepSindhu Ahh, well that's an awkward definition. The point is that the energy might be stored/carried in the object/system. But to release it, a force has to appear. And we define work directly based on this force: $$W=\int F\cdot dr$$ We might "say" it as if it is the object because that's a simple way to talk about it, but that doesn't make much sense physically. $\endgroup$
    – Steeven
    Jun 25, 2020 at 11:08
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    $\begingroup$ Thank You, that clears my understanding of work and energy, for now $\endgroup$
    – Rajdeep Sindhu
    Jun 25, 2020 at 11:14
  • $\begingroup$ I just faced a contradiction while assuming that the ball does not do any work. it was while I was deriving the formula for Kinetic Energy. The method of derivation that my textbook tells is : Assume that an object is moving with a velocity of $v$ and a force $F$ is acting in the opposite direction. The amount of work done by the object till it comes to rest will be the kinetic energy possessed by it. Now, if we assume that $W = 0$, we would get $KE = 0$, which is not the case. Instead, if we assume that the ball does some work, we do obtain the right answer (maybe) ..... $\endgroup$
    – Rajdeep Sindhu
    Jun 25, 2020 at 13:07
  • $\begingroup$ ..... Let mass of the body be $m$. So, acceleration = $\dfrac{F}{m}$. By third equation of motion, $v^2=u^2+2as \implies s = \dfrac{v^2-u^2}{2a} = \dfrac{0^2-v^2}{\dfrac{2F}{m}} = \dfrac{-mv^2}{2F}$. Now, by Newton's third law of motion, force exerted by ball against the opposing force is $-F$. Hence, $W = Fs = (-F).\dfrac{-mv^2}{2F} = \dfrac{mv^2}{2}$. Something about this feels off to me as well. Like, what is the ball doing work on? How would we derive the expression for Kinetic Energy assuming that the ball doesn't do any work? Thanks! $\endgroup$
    – Rajdeep Sindhu
    Jun 25, 2020 at 13:11

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