2
$\begingroup$

Consider two observers moving with respect to each other at a speed $v$ along a straight line. They observe a block of mass $m$ moving a distance $l$ on a rough surface. The change in KE of the block is different to the different observers in this case. Here "work done by friction= change in KE" i.e. it is different to different observers. How to solve the same using "work done by friction= -μmgl"? It appears to me that by the 2nd way the answer is coming different. As μ, m, g and l are constants, the work done by friction will be the same to the different observers. Please resolve the controversy.

$\endgroup$

1 Answer 1

3
$\begingroup$

Actually, $l$ is not a constant. The different observers measure different values for $l$. The discrepancy in the change in KE is entirely explained by this disagreement on $l$.

So, for a block undergoing a frictional force of $-\mu m g$ starting at $x(0)=0$ with an initial velocity of $v(0)=\dot x(0)=v_0$ we get $$x(t)=-\frac{1}{2}\mu g t^2 + v_0 t$$ so at every point in time the change in KE is $$\Delta KE = \frac{1}{2} m \left(\dot x (t)^2-\dot x(0)^2\right) = \frac{1}{2}\mu^2 m g^2 t^2-\mu m g v_0 t$$ and the other formula is $$ -\mu m g l = - \mu m g x(t)=\frac{1}{2}\mu^2 m g^2 t^2-\mu m g v_0 t$$ which is always equal to the change in KE.

$\endgroup$
2
  • $\begingroup$ There's an opportunity to make this answer even clearer and more useful by considering the paths and apparent changes in velocity for a friction-braked 1 kg block that appears to transition from 3 m/s to 1 m/s, from 2 m/s to 0 m/s (in the frame of the rough surface), and 1 m/s to -1 m/s, for instance. $\endgroup$ May 24 at 19:48
  • $\begingroup$ @Chemomechanics good idea, but I will do it symbolically instead of for a specific numerical example $\endgroup$
    – Dale
    May 24 at 20:27

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.