A block of mass $5 \mathrm{~kg}$ moving on a rough surface with a velocity of $4 \mathrm{~ms}^{-1}$ is stopped by the friction in 2 seconds. Then the coefficient of friction between the contact surfaces is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Question

Aicharya · Accepted Answer

Mass of block, $\mathrm{m}=5 \mathrm{~kg}$ velocity $\mathrm{v}=4 \mathrm{~m} / \mathrm{s}$ $\begin{aligned} & \mathrm{F}=\mu \mathrm{mg} \Rightarrow \mathrm{ma}=\mu \mathrm{mg} \ & \frac{\mathrm{mv}}{\mathrm{t}}=\mu \mathrm{mg} \Rightarrow \mu=\frac{\mathrm{v}}{\mathrm{gt}}=\frac{4}{2 	imes 10}=0.2 \end{aligned}$

A block of mass $5 \mathrm{~kg}$ moving on a rough surface with a velocity of $4 \mathrm{~ms}^{-1}$ is…

Solution