An inclined plane has a rough and smooth surface which are of equal length. A ball is kept at rest at a height $h$ and is then rolled on the surface of the inclined plane.

My question is that shouldn't the work done by friction when the rotating object first moves through the rough surface and then to the smooth surface be same as the work done by friction when the rotating object first passes through the smooth surface and then to the rough surface. (All other conditions are the same in each case.)

My textbook tells that work done by friction in the second case is greater. Can someone explain where I am going wrong?

  • $\begingroup$ How is the ball rolling on the incline? Is it released from rest at the top? Is it launched from the bottom? Is it different for each example? Is the incline the same in either case, or do the rough/smooth parts switch between examples? More information is needed here to answer the question. $\endgroup$ Jan 23, 2019 at 14:29
  • $\begingroup$ Work is done by the friction force if the ball is slipping/sliding. $\endgroup$
    – Gert
    Jan 23, 2019 at 14:56
  • $\begingroup$ @Gert Static friction can still do work $\endgroup$ Jan 23, 2019 at 15:05
  • $\begingroup$ lamdeb, I am still confused. Are the rough/smooth patches changing position between examples? And from your wording it seems like you are saying someone applies a force to get the ball rolling before let go. Is this the case? $\endgroup$ Jan 23, 2019 at 15:07
  • $\begingroup$ i am saying that there are two identical inclined planes but in the first incline plane the the first half of its slope is a rough surface and the other half is smooth surface whereas in the second inclined plane the first half is smooth surface and the second half is rough.surface. No force is applied on the ball , it is just kept at the top of the inclined plane and left so that it rolls down. $\endgroup$
    – lamdeb
    Jan 23, 2019 at 15:32

2 Answers 2


In spite of missing details, I will do my best to explain what I think is going on here. I will edit my answer if more details are given.

The only way I can think that there is no work done when going from rough to smooth and there is work going from smooth to rough is assuming two things (clarify if I am wrong)

1) The rough/smooth patches are flipped between examples. So in the first case the rough patch is at the top, and in the second case the smooth part is at the top

2) The question is actually talking about either "work done by kinetic friction" or "energy dissipated by friction".

If the above are correct, then an explanation becomes pretty evident: In the first case, since the ball starts at rest, static friction will cause the ball to roll without slipping. Because there is no slipping, there is no kinetic friction to dissipate energy. However, in the second case when the ball hits the rough patch it has translational motion without rotating. Therefore, when the ball hits the rough patch it will start rolling while slipping as kinetic friction acts on the ball until it can roll without slipping (assuming the incline is long enough).

So in the first case no energy is dissipated due to friction, but in the second case energy is dissipated by friction.


Your book (most likely) takes into consideration the speed the ball reaches while on the smooth surface (in second example) and depending on the rough surface, the friction may depend on this speed (like air resistance is greater at greater speeds)

  • $\begingroup$ Yes,my book asks which case does the rotationg object have maximun kinetic energy at the bottom of the inclined plane , which is linked to the work done by friction. My book tells that in the first case work done zero whereas in the second case work done is non zero. I am not able to understand this. $\endgroup$
    – lamdeb
    Jan 23, 2019 at 15:04
  • $\begingroup$ I don't see how work done in first case can be zero either. Even in low speed, ball going through rough surface would meet some resistance. Does your book specify what kind of rough surface that is? $\endgroup$
    – Alex Doe
    Jan 23, 2019 at 15:16
  • $\begingroup$ no it does not . It just said "rough surface" $\endgroup$
    – lamdeb
    Jan 23, 2019 at 15:20
  • $\begingroup$ Alex, I think there are some details in the set up being left out. Hopefully the OP can clarify things in the question. $\endgroup$ Jan 23, 2019 at 15:21

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