A block of mass 10 kg is moving with constant velocity $10m/s$ in rightward direction on a rough horizontal surface with a force of $50N$ acting on it.Calculate the work done by dynamic friction on the block and net work done by friction on ground for 5 s

Attempt: Now since the block is moving with constant velocity hence friction will be equal to the force ie $50N$.

So work done by the friction on block will be $-2500J$.

For the ground, net work by friction also should be $-2500J$, because friction, acting on the ground, will be in forward direction, but the ground will appear to move backwards relative to the block, but the book gives this value as $2500J$. What am I doing wrong?

$$mv_0^2=8mgh\implies v_0=2\sqrt{2gh}$$ $$\frac{2}{3}v_0=\frac{4}{3}\sqrt{2gh}$$ My answer is $$2\sqrt{\frac{2}{3}gh}$$


closed as off-topic by Emilio Pisanty, sammy gerbil, rob Jul 30 '18 at 0:08

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    $\begingroup$ The units of work should be J(oules). $\endgroup$ – V.F. Jul 29 '18 at 13:47

The work performed by the friction force applied by the block to the ground, is not measured by how much the block has moved the ground as a whole, but rather by how much it has moved/deformed microparticles/microstructures of the rough surfaces at the interface between the block and the ground.

All this movements and deformations will be in the forward direction, i.e., in the direction of the friction force applied by the moving block to the ground, and, therefore, the work, performed by the block, will be positive.

Strictly speaking, we cannot even say that all of this work is applied to the ground surface, since the bottom surface of the block will receive a fair share of this work as well, heating up and wearing in the process.


Consider the problem from the following perspective. Let the block be the system. Then the work done on the block by the external 50N force is


The work is negative because it is done on the system.

Since the velocity is constant during the 5 seconds, there is no change in kinetic energy. If there were no friction there would be a change in kinetic energy per the work energy principle:

$Fd=\Delta KE$

That means the work done by the block against friction must be +2500J so that the total work done on the block during the 5 s is zero resulting in no change in kinetic energy. The work against friction raises the temperatures of both surfaces, like vigorously rubbing your hands together.

Does this make sense?