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(A) $1$. The crate is dropped vertically,so its initial horizontal velocity is $0$. The belt moves at a constant velocity $v = 1.20\ m/s$.$2$. The for

Vedclass · Accepted Answer

$432$

Vedclass · Answer

(A) $1$. The crate is dropped vertically,so its initial horizontal velocity is $0$. The belt moves at a constant velocity $v = 1.20\ m/s$.$2$. The force of kinetic friction acting on the crate is $f_k = \mu_k N = \mu_k mg = 0.400 	imes 300\ kg 	imes 9.8\ m/s^2 = 1176\ N$.$3$. The crate accelerates until its velocity matches the belt's velocity $v = 1.20\ m/s$. The time taken is $t = v/a$,where $a = f_k/m = \mu_k g = 0.400 	imes 9.8 = 3.92\ m/s^2$. So,$t = 1.20 / 3.92 \approx 0.306\ s$.$4$. The displacement of the crate during this time is $s_c = \frac{1}{2}at^2 = \frac{1}{2} 	imes 3.92 	imes (0.306)^2 \approx 0.1837\ m$.$5$. The displacement of the belt during this time is $s_b = v 	imes t = 1.20 	imes 0.306 \approx 0.3673\ m$.$6$. The work done by the motor must overcome the friction force exerted by the crate on the belt. The force exerted by the crate on the belt is $f_k$ in the direction of the belt's motion. The motor must exert an equal and opposite force to maintain constant speed. Thus,$W_{motor} = f_k 	imes s_b = 1176\ N 	imes 0.3673\ m \approx 432\ J$.

Column-$I$	Column-$II$
$(1)$ Static friction	$(a)$ Limiting friction
$(2)$ Rolling friction	$(b)$ Ball bearing
	$(c)$ Object moving on a road

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