Question

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

The block is released from rest with the spring in the unstretched position. The block moves $ 10\, cm $ down the incline before coming to rest. The coefficient of friction between the block and the incline is
(Take $ g = 10\, ms^{-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 = 1\, kg $ , $ \theta = 45^° $ , $ k = 100 \,N \,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 = 10\, cm = 10 \times 10^{-2}\,m $
In equilibrium,
Work done = Potential energy of stretched spring
$ mg(sin\theta - \mu cos\theta)x = \frac{1}{2}kx^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} $
$ 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 $