A railroad car moves under a grain elevator at a constant speed of 3.70 m/s. Grain drops into the car at the rate of 580 kg/min. What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?
F = Ma -> R * V(rel)
Sorry - got cut off. I meant to supply
R * V(rel) = Ma
To find the magnitude of the force needed to keep the railroad car moving at a constant speed, we need to determine the net force acting on the car.
Given information:
- Speed of the car (v) = 3.70 m/s
- Rate at which grain drops into the car (mass flow rate, dm/dt) = 580 kg/min
First, let's convert the mass flow rate from kg/min to kg/s:
Mass flow rate (dm/dt) = 580 kg/min = (580 kg/min) * (1 min/60 s) = 9.67 kg/s
Next, we can calculate the force needed by using Newton's second law (F = ma) and considering the conservation of momentum principle.
According to the conservation of momentum principle, the rate of change of momentum is equal to the product of mass and acceleration (F = dm/dt * v(rel)), where v(rel) is the relative velocity between the car and the grain.
In this case, the car's speed is constant, so the relative velocity between the car and the grain is also 3.70 m/s.
Therefore, the magnitude of the force needed to keep the car moving at constant speed is:
F = (dm/dt) * v(rel)
F = (9.67 kg/s) * (3.70 m/s)
F ≈ 35.77 N
Hence, the magnitude of the force needed to keep the car moving at constant speed, assuming negligible friction, is approximately 35.77 Newtons.