Question

A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between the table and the block is 0⋅1, how far does the block move before coming to rest?

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=250\mathrm{gm}=0.250\mathrm{kg} \\ \mathrm{Initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block},u=40\mathrm{cm}/s=0.4\mathrm{m}/s \\ \mathrm{Final}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{block},\nu =0 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{friction},\mu =0.1 \end{gathered}$$

Force in the forward direction is equal to the friction force.
$$\begin{gathered} \mathrm{Here},\mu \mathrm{R}=ma \\ \left(\mathrm{where}a\mathrm{is}\mathrm{deceleration}\right) \\ a=\left(\frac{\mu \mathrm{R}}{m}\right)=\left(\frac{\mu mg}{m}\right)=\mu g \\ =0.1\times 9.8=0.98m/s^2 \\ s=\frac{{\nu }^2-u^2}{2a}=0.082\mathrm{m} \\ =8.2\mathrm{cm} \end{gathered}$$

Again, work done against friction,

$$\begin{gathered} \mathrm{W}=-\mu Rs\cos \mathrm{\theta } \\ =-1\times 2.5\times 0.082\times 1 \\ =-0.02\mathrm{J} \\ \Rightarrow \mathrm{W}=-0.02\mathrm{J} \end{gathered}$$