Conveyor belt

I have 2 methods, i would state the 2nd one, which is very simple and conceptual.

Concept: Whatever the work belt did, it must have gone somewhere, where?? Ans: To increase the kinetic energy of the block and some lost in friction, while the time the block was slipping

In ground frame the energy gained is the energy the block attains finally. In the frame of belt friction does its work on the block, the work energy equivalence holds so in belt frame the initial velocity of the block in not zero(find the relative velocity) initially but finally it gets to zero(same as that of the belt so relative zero), the work done is used up in taking away that energy..

We can work out the amount of energy involved in each case 1. In the ground frame, the block starts at rest and eventually moves at speed v. The kinetic energy gained is 1/2 m(v^2) 2. By moving to the frame of the conveyor belt, we can find the amount of energy lost to friction. In that frame, the block originally has energy v and then slows down to rest, with all the original energy dissipated to friction. In total, the energy lost to friction is 1/2 m (v^2) The total work down by the belt is then 2* 1/2 *m *(v^2) 432J

I solved it this way. Though it has been a long time since I had solved some mechanics, so I don't know whether my solution is logically right or wrong. I leave it to you to judge it.

When the block touches the bed, the frictional force acting on the block will be f=300 10 0.4=1200; Now we don't know the time for which the slipping took place. Therefore, by Impulse-momentum principle we have f t=m v-m u here f=1200N,u=0, m=300 and v=1.2; therefore t=0.3 sec; Now power of the motor will be P= F.v=1200 1.2=1440 W Therefore, work of the motor will be W=P . t = 1440*0.3 = 432J

The acceleration of the block is $\mu g$ . Since ${v}_{f} = {v}_{0} + at$ , we solve for $t$ . The distance travelled is equal the velocity of the belt times time: $d = \frac{{{v}_{belt}}^{2}}{\mu g}.$ $W = {F}_{f} \cdot d$ $W = \mu mg \frac{{{v}_{belt}}^{2}}{\mu g}$ $W = m {{v}_{belt}}^{2}$ Plug in the values to get 432 Joules.