# Find the extra force required to maintain the speed of the belt

1. Sep 16, 2010

### Doubell

1. The problem statement, all variables and given/known data

coal is falling onto a conveyor belt at a rate of 540 tonnes every hour. the belt is moving horizontally at 2.0 ms-1. find the extra force required to maintain the speed of the belt. (1 tonne=1000kg.)

2. Relevant equations

F=m*a

3. The attempt at a solution

i tried to solve by saying that F=m*a. so to find force i would need the accelaration of the belt and multiply it by the mass of the coal. the acceleration is the rate of change of velocity so i would use the 2ms-1 as the final velocity and 0 ms-1 as the annitial subtracting 0 from two leaves 2 ms-1 and i divide by the rate of change of time which is 3600s.this leaves me with an accelaeration of 0.000556 ms/s^2 multiplying this by the mass of the coal 540000kg leaves 300 N. is this right

2. Sep 16, 2010

### vandersmissen

Re: Mechanics

I'm not too sure on that problem. What I think I can advise however is a few things.

You are trying to maintain the speed, this means that
final velocity and initial velocity are both 2 m/s which means that

v-v_0=at
2m/s - 2m/s = at
0=at

so there is no acceleration.

If you take the 540,000kg/h and convert it into seconds you get 150kg/s . So now you know how much the mass is increasing by per second and the acceleration.

Using dimensional analysis, we have 150 kg/s and 2.0 m/s , if we multiply these values, we get 300 kg.m/s^2 which is a newton , so I believe the answer is 300 N like you suspected. However I would also check with others as I am not certain on this.

3. Sep 16, 2010

### Doubell

Re: Mechanics

i just looked at it and found out another solution. in accordance to newton's second law force = rate of change of momentum. the innitial momentum that occured in 1 second would be 150kg *2 ms-1 (mass x velocity) which equals 300 kg ms-1 the final momentum that occured in 3600 seconds would be 540000kg *2 ms-1.( the final and innitial velocities are the same) which = 1080000 kg ms-1. change in momentum = final momentum - innitial momentum this equals 1079700 kg ms-1. the change in time = T2-T1 = 3600s-1s which gives 3599s. Now in accordance to Newtons second law the rate of change of momentum = force this implies that change in momentum(1079700 kg ms-1) divided by change in time(3599s) = 300 kg m/s^2 =300N