Question

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min. What is the magnitude of the force needed to keep the car moving constant speed if friction is negligible?

Answer 1

F = 768 N

Step-by-step explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min,
`(dm)/(dt)=240\ kg/min = 4\ kg/s`

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

`F=(dp)/(dt)`

`F=m(dv)/(dt)+v(dm)/(dt)`

Since, the speed is constant,

`F=m(dv)/(dt)`

`F=v(dm)/(dt)`

`F=3.2* 240`

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.