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)

$0.3$

Answer 2

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 $mg(sin\theta - \mu cos\theta)x = \frac{1}{2}kx^2$ $2mg \,(sin\theta - \mu cos\theta) = kx$ $2 \times 1 \times 10 \times \left(sin45^{\circ}-\mu cos45^{\circ}\right)$ $= 100 \times 10 \times 10^{-2}$ $sin\,45^{\circ} - \mu\,cos\,45^{\circ} = \frac{1}{2}$ $\frac{1}{\sqrt{2}}-\frac{\mu}{\sqrt{2}} = \frac{1}{2}$ $1-\mu = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ $\Rightarrow \mu = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$ $\mu = 0.3$