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

The Correct Option is 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 $