Question

A block of mass $1\text{kg}$ is situated on a rough inclined plane and is connected to a spring of spring constant $100{\text{Nm}}^{-1}$ as shown in figure. The block is released from rest when the spring is unstretched. The block moves through a distance of $10\text{cm}$ down the plane before coming to rest. Assuming that spring and pulley are massless, and the pulley is frictionless. Find the coefficient of friction between the block and the inclined surface.
(Take $g=10{\text{ms}}^{-2}$)

• A. $\frac{1}{6}$
• B. $\frac{1}{8}$
• C. $\frac{1}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is B $\frac{1}{8}$

Given:
$m=1\text{mg};k=100\text{N/m};x=10\text{cm}$


Considering extension in the spring as $x$, using work-energy theorem,

$W_{\text{con force}}+W_{\text{non -con force}}+W_{\text{ext}}=\Delta K$

$mgh-f_{s}x-\frac{1}{2}k{x}^2=0[\because \text{The block starts from rest and comes to rest}]$

$mg\left(x\sin {37}^{\circ }\right)=\frac{1}{2}k{x}^2+\left(\mu mg\cos {37}^{\circ }\right)x$

$\Rightarrow 1\times 10\times 0.1\times \frac{3}{5}=[\frac{1}{2}\times 100\times (0.1{)}^2]+(\mu \times 1\times 10\times \frac{4}{5}\times 0.1)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+(\frac{4}{5}\times \mu )$

$\therefore \mu =\frac{1}{8}=0.125$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.