# What is the work done by friction on a body from the perspective of different observers using the formula $-μmgl$?

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.

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.