Don't drop it!

When the crate of mass $m=300$ kg is dropped and touches the belt, its initial horizontal velocity $u=0$ m/s. It slides and accelerates to the belt velocity $v=1.20$ m/s after a short time $t$ s and making a displacement of $s$ m.

The crate is accelerated by the frictional force $F=\mu mg$ N, where $\mu=0.400$ is the coefficient of friction between the crate and belt, and $g=9.8$ $m/s^2$ is the acceleration due to gravity. $\Rightarrow F = 0.400 \times 300 \times 9.8 = 1176$ N.

Since $F=ma$ , the acceleration $a=\frac {F}{m} = \frac {1176}{300} = 3.92$ $m/s^2$ .

The time taken to accelerate $t=\frac {v}{a}=\frac {1.20}{3.92}= 0.306122449$ s.

The displacement made $s=\frac {1}{2}at^2=0.183673469$ m.

The work $W$ done by the motor is to overcome the fictional force and accelerate the 300kg crate. That is: $W = (F+ma)s = 0.183673469(1176+300 \times 3.92)$ $=\boxed{432}$ J.

I used work energy theorem and got the answer in a single step...

Acc of block =0.4g. v=u+at

1.2=4t

t=0.3sec . distance moved by belt=1.2t

1.2*0.3=0.36m

now on mass 300

acc=4

m=300

W=F.s (1-D)

W=m a s

W=300 4 .36

W=432

Dont forget to follow me.

E=mechanical energy , Eth= thermal energy and Ei=internal energy

E2=E1-Eth-Ei......................................1

E1=0 because the initial kinetic and potential energy of system is zero.

then after some time it gain velocity 1.20 m/s so.

0.5 x m x v^2 = -Eth because the work done by frictional force

216 = -Eth

therefore E2= -Eth

now taking eq 1

216 = -216 - Ei

216x2 = Ei

432 = Ei therfore work done by motor is 432