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?

Respuesta :

Answer:

F = 768 N                  

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, [tex]\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s[/tex]

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 :

[tex]F=\dfrac{dp}{dt}[/tex]

[tex]F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}[/tex]

Since, the speed is constant,

[tex]F=m\dfrac{dv}{dt}[/tex]

[tex]F=v\dfrac{dm}{dt}[/tex]

[tex]F=3.2\times 240[/tex]

F = 768 N

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

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