A 1; kg block situated on a rough incline is | vedclass

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(C) Mass of the block,$m = 1\; kg$. Spring constant,$k = 100\; N m^{-1}$. Displacement of the block,$x = 10\; cm = 0.1\; m$. According to the work-ene

Vedclass · Accepted Answer

$0.115$

Vedclass · Answer

(C) Mass of the block,$m = 1\; kg$. Spring constant,$k = 100\; N m^{-1}$. Displacement of the block,$x = 10\; cm = 0.1\; m$. According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy. Since the block starts from rest and comes to rest,the change in kinetic energy is zero. Work done by gravity + Work done by spring + Work done by friction = $0$. $mgx \sin 37^{\circ} - \frac{1}{2} k x^2 - f x = 0$. Here,$f = \mu N = \mu mg \cos 37^{\circ}$. Substituting the values: $(1)(9.8)(0.1) \sin 37^{\circ} - \frac{1}{2} (100)(0.1)^2 - \mu (1)(9.8)(0.1) \cos 37^{\circ} = 0$. $0.98 	imes 0.6018 - 0.5 - \mu (0.98)(0.7986) = 0$. $0.5898 - 0.5 = \mu (0.7826)$. $0.0898 = \mu (0.7826)$. $\mu = \frac{0.0898}{0.7826} \approx 0.115$.

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