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(C) Let $m = 0.1 \,kg$, $\mu = 0.2$, $g = 10 \,ms^{-2}$, $	heta = 30^{\circ}$, and $s_2 = 1 \,m$ (horizontal distance).On the horizontal plane, the r

Vedclass · Accepted Answer

$0.306$

Vedclass · Answer

(C) Let $m = 0.1 \,kg$, $\mu = 0.2$, $g = 10 \,ms^{-2}$, $	heta = 30^{\circ}$, and $s_2 = 1 \,m$ (horizontal distance).On the horizontal plane, the retardation $a_2 = \mu g = 0.2 	imes 10 = 2 \,ms^{-2}$.Using $v^2 = u^2 - 2a_2s_2$, where $v=0$ (comes to rest), we get $0 = u^2 - 2(2)(1)$, so $u^2 = 4$, or $u = 2 \,ms^{-1}$. This is the velocity at the bottom of the incline.On the inclined plane, the net acceleration $a_1 = g(\sin 	heta - \mu \cos 	heta) = 10(\sin 30^{\circ} - 0.2 \cos 30^{\circ}) = 10(0.5 - 0.2 	imes 0.866) = 10(0.5 - 0.1732) = 3.268 \,ms^{-2}$.Using $v^2 = u_0^2 + 2a_1s_1$, where $u_0=0$ and $v=2 \,ms^{-1}$, we get $4 = 2(3.268)s_1$, so $s_1 = 4 / 6.536 \approx 0.612 \,m$.The work done by friction on the incline is $W_1 = -f_1 s_1 = -(\mu mg \cos 30^{\circ}) s_1 = -(0.2 	imes 0.1 	imes 10 	imes 0.866) 	imes 0.612 \approx -0.106 \,J$.The work done by friction on the horizontal plane is $W_2 = -f_2 s_2 = -(\mu mg) s_2 = -(0.2 	imes 0.1 	imes 10) 	imes 1 = -0.2 \,J$.The total work done by friction is $W = W_1 + W_2 = -0.106 - 0.2 = -0.306 \,J$. The magnitude is $0.306 \,J$.

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