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Work, Power and Energy

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Question : 26 of 30
Marks: +1, -0
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m−1100\,\mathrm{N\,m}^{-1} as shown in figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before comingtorest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.
Solution:  
As is clear from figure.
R=mgcos⁡θR = mg \cos\theta
F=μR=μmgcos⁡θF = \mu R = \mu mg \cos\theta1Net force on the block down the incline
=mgsin⁡θ−F= mg\sin\theta - F
=mgsin⁡θ−μmgcos⁡θ= mg\sin\theta - \mu mg \cos\theta =mg(sin⁡θ−μcos⁡θ)= mg(\sin\theta - \mu \cos\theta)
Distance moved, x=10 cm=0.1 mx = 10\,\text{cm} = 0.1\,\text{m}.
In equilibrium, work done = P.E. of stretched spring
mg(sin⁡θ−μcos⁡θ)x=12kx2mg(\sin\theta - \mu \cos\theta) x = \frac{1}{2} k x^{2}
2mg(sin⁡θ−μcos⁡θ)=kx2mg(\sin\theta - \mu \cos\theta) = kx
2×1×10(sin⁡37∘−μcos⁡37∘)=100×0.12 \times 1 \times 10 (\sin 37^{\circ} - \mu \cos 37^{\circ}) = 100 \times 0.1
20(0.601−μ⋅0.798)=10∴μ=0.12620(0.601 - \mu \cdot 0.798) = 10 \therefore \mu = 0.126
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