3

My teacher told me that friction can have positive work which is true. But he told that a pair of friction forces can never have positive work. I am not able to think the reason for this statement. Any help will be really appreciated.

Abhay Sharma
  • 39
  • 1
    @user1936752 That is inaccurate. Friction is in the opposite direction of relative motion between two surfaces, but it may be in the same direction as the velocity as measured in a specific reference frame. – Chris Nov 23 '18 at 12:59
  • Ah yes, I'll delete the misleading comment! – user1936752 Nov 23 '18 at 15:26

2 Answers2

3

Farcher's answer covers the static friction case, but does not handle the situation in general. I will try to do this here.

Let's say we have one object on top of another object, and both are moving at a constant speed. Then, we apply a force $F$ to one of the blocks. Without loss of generality, let's say we push on the top block. The resultant relative velocity between the blocks is $w$. This set up covers the case of static friction ($w=0$) as well as the case of kinetic friction ($w\neq0$ and might not be constant).

Now, it will be easier to look at the instantaneous power delivered to each block by friction given by $$P=\mathbf f\cdot\mathbf v_b$$ where $\mathbf f$ is the friction force, and $\mathbf v_b$ is the velocity of the block at some point in time.

Without loss of generality let's assume the blocks were originally moving to the right, and the applied force on the top block is also to the right. Then at some point in time, the velocity of the bottom block will be $v$, and the velocity of the top block will be $v+w$. The friction force on the top block is to the left, and the friction force on the bottom block is to the right. By Newton's third law, each friction force has the same magnitude $f$.

On the top block: $$P_T=-f(v+w)$$

On the bottom block: $$P_B=fv$$

Therefore, the instantaneous net power due to friction in the system is given by $$P=P_T+P_B=-f(v+w)+fv=-fw\leq0$$

Now, as the blocks accelerate, $v$ and $w$ might be changing, but the signs will not change, so this result holds for any scenario.

Therefore, the work done by a friction pair cannot be positive.

Also note that, even though the work done by a single friction force depends on the reference frame (as you can see above, it can actually be positive), the net power does not depend on the reference frame. It only depends on the relative velocity between the blocks. $-fw$ gives the rate at which energy is dissipated in our system.

BioPhysicist
  • 56,248
  • 1
    Good approach. I like this way of analyzing the work done by friction. It is especially useful for questions related to friction in different reference frames – Dale Nov 29 '18 at 03:38
  • 1
    @Dale to be honest I must give you some credit, since I generalized what I learned from your approach here: https://physics.stackexchange.com/questions/442074/friction-and-work-in-inertial-reference-frames – BioPhysicist Nov 29 '18 at 03:43
2

Imagine one block $T$ on top of another block $B$ with both blocks moving with a velocity $\vec v$ to the right ie not moving relative to one another.
A force $\vec F$ is being applied to the bottom block $B$ to cause an acceleration of both blocks to the right.

The friction force on the top block due to the bottom block is $\vec F_{\rm TB}$ and this force is in the direction of motion of the top block (to the right) so the work done by that force is positive.

The friction force on the bottom block due to the top block is $\vec F_{\rm BT} (= -\vec F_{\rm TB}$ - Newton's thrid law) and this force is in the opposite direction to the motion of the top block so the work done by that force is negative.

In short, the displacement of both blocks is in the same direction but the frictional forces which are applied to the two blocks are in opposite directions, so the work done by friction on one block is positive whilst the work done by friction on the other block is negative.
Overall the work done by the frictional forces is zero.

Farcher
  • 95,680