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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 fo

Vedclass · Accepted Answer

$432$

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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 force of friction acting on the crate is $f = \mu mg = 0.400 	imes 300 	imes 9.8 = 1176 \ N$.$3$. The acceleration of the crate is $a = f/m = \mu g = 0.400 	imes 9.8 = 3.92 \ m/s^2$.$4$. The time taken for the crate to reach the belt's speed is $t = v/a = 1.20 / 3.92 \approx 0.306 \ s$.$5$. The distance moved by the belt during this time is $d_{belt} = v 	imes t = 1.20 	imes (1.20 / 3.92) = 1.44 / 3.92 \approx 0.367 \ m$.$6$. The motor must exert a force equal to the friction force to maintain constant speed,so $F_{motor} = f = 1176 \ N$.$7$. The work done by the motor is $W = F_{motor} 	imes d_{belt} = 1176 	imes (1.44 / 3.92) = 432 \ J$.

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