Question

A $1kg$ block situated on a rough incline is connected to a spring of negligible mass having spring constant $100N{m}^{-1}$ as shown in the figure.

The block is released from rest with the spring in the unstretched position. The block moves $10cm$ down the incline before coming to rest. The coefficient of friction between the block and the incline is
(Take $g=10m{s}^{-2}$ and assume that the pulley is frictionless)

• A. $0.2$
• B. $0.3$
• C. $0.5$
• D. $0.6$

Answer 1

Answer: (B)

Here, $m=1kg$ , $\theta =45^{°}$ , $k=100N{m}^{-1}$
From figure, $N=mgcos\theta$
$f=\mu N=\mu mgcos\theta$
where $\mu$ is the coefficient of friction between the block and the incline.
Net force on the block down the incline,
$=mgsin\theta -f$
$=mgsin\theta -\mu mgcos\theta =mg(sin\theta -\mu cos\theta )$
Distance moved, $x=10cm=10\times 10^{-2}m$
In equilibrium,
Work done = Potential energy of stretched spring
$mg(sin\theta -\mu cos\theta )x=\frac{1}{2}k{x}^2$
$2mg(sin\theta -\mu cos\theta )=kx$
$2\times 1\times 10\times \left(sin45^{\circ }-\mu cos45^{\circ }\right)$
$=100\times 10\times 10^{-2}$
$sin45^{\circ }-\mu cos45^{\circ }=\frac{1}{2}$
$\frac{1}{\sqrt{2}}-\frac{\mu }{\sqrt{2}}=\frac{1}{2}$
$1-\mu =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \mu =1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$
$\mu =0.3$