Physics - NLM Question with Solution | TestHub

The forces acting ont he block are shown in figure. here $\mathrm{M}=2.5\mathrm{kg}$ and $\mathrm{F}=15\mathrm{N}$. When $\mathrm{F}=15\mathrm{N}$ is appplied to the block, the block remains inlimiting equilibrium. The force of friction is thus $\mathrm{f}={\mu }_{\mathrm{s}}\mathrm{N}$. Applying Newton's first law, $\mathrm{f}=\mu \mathrm{N}\text{ and }\mathrm{N}=\mathrm{mg}$ so that $\mathrm{F}={\mu }_{\mathrm{s}}\mathrm{Mg}$

or, ${\mu }_{\mathrm{s}}=\frac{\mathrm{F}}{\mathrm{mg}}=\frac{15\mathrm{N}}{(2.5\mathrm{kg})\left(10\mathrm{m}/{\mathrm{s}}^2\right)}=0.60$

When the block is gently pushed to start the motion, kinetic friction acts between the block and the surface. Since the block takes 5 seconds to slide through the first 10 m, the acceleration a is given by $10\mathrm{m}=\frac{1}{2}\mathrm{a}(5\mathrm{s}{)}^2$

or, $\mathrm{a}=\frac{20}{25}\mathrm{m}/{\mathrm{s}}^2=0.8\mathrm{m}/{\mathrm{s}}^2$

The frictional force is $\mathrm{f}={\mu }_{\mathrm{k}}\mathrm{N}={\mu }_{\mathrm{k}}\mathrm{Mg}$ Applying Newton's second law $\mathrm{F}-{\mu }_{\mathrm{k}}\mathrm{Mg}=\mathrm{Ma}$

or, ${\mu }_{\mathrm{k}}=\frac{\mathrm{F}-\mathrm{Ma}}{\mathrm{Mg}}=\frac{15\mathrm{N}-(2.5\mathrm{kg})\left(0.8\mathrm{m}/{\mathrm{s}}^2\right)}{(2.5\mathrm{kg})\left(10\mathrm{m}/{\mathrm{s}}^2\right)}=0.52$