KVPY SA Exam 2014 Question Paper

When box is dropped from a height h, then speed at ground is v, therefore using mechanical energy conservation
$\mathrm{mgh}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2$ …(i)
when body slides on rough inclined plane, friction force will also act f = μN = μ mg cos θ
Applying work-energy theorem
$\mathrm{mgh}\mathit{–}\mathrm{fs}=\frac{1}{2}\mathit{m}{\left(\frac{\mathit{v}}{3}\right)}^2-0$
$\mathrm{mgh}\mathit{–}\mathrm{f}\frac{\mathit{h}}{\sin \mathit{\theta }}=\frac{1}{2}\mathit{m}{\left(\frac{\mathit{v}}{3}\right)}^2$
$\left(\sin \mathit{\theta }=\frac{\mathit{h}}{\mathit{s}}\right)$
$\mathrm{mgh}-\mathrm{\mu }\mathrm{mg}\cos \mathit{\theta }\times \frac{\mathit{h}}{\sin \mathit{\theta }}=\frac{1}{2}\frac{{\mathit{m}\mathit{v}}^2}{9}$ …(ii)
from equation (i) & (ii)
$\mathrm{mgh}\left[1-\mathrm{\mu }\cot \mathit{\theta }\right]=\left(\frac{1}{9}\right)\mathrm{mgh}$
putting θ = 45º, cotθ = 1
$1-\mathrm{\mu }=\frac{1}{9}$
⇒ $\mathrm{\mu }=\frac{8}{9}$