Question

A 5 kg Blocks slides along a rough horizontal surface the coefficient of kinetic friction for the block and the surface is 0.4 at what rate in watts is the frictional force doing work on the block at the instant when it’s Speed s 5 m/s

Answer 1

By definition, the work rate is:

$$P = F_{fr}v$$

where $v = 5m/s$ and $F_{fr}$ is the frictional force. In order to find it one can note, that the frictional force is equat to the normal force $N$ times the coefficient of kinetic friction $\mu = 0.4$ :

$$F_{fr} = \mu N$$

On a horizontal surface the normal force is equal to the force of gravity:

$$N = mg$$

where $m = 5kg$ and $g = 9.81 m/s^2$.

Combining it all together, obtain

$$P = F_{fr}v = \mu mgv = 0.4\cdot 5\cdot 9.81\cdot 5 = 98.1\space W$$

Answer. 98.1 W.