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

The correct option is B $0.3$

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