A loaded ore car has a mass of 950 kg and rolls on rails with negligible frictio

Question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 33.0° above the horizontal. The car accelerates uniformly to a speed of 2.05 m/s in 12.5 s and then continues at constant speed (a) What power must the winch motor provide when the car is moving at constant speed? Enter a number hat must be exerted by the cable in order to keep the cart moving at constant velocity and use that force to find the power required. kW (b) What maximum power must the motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1,250 m? Need Help? Read ItMaster Submit Answer Save Progress

Explanation / Answer

Power is the rate at which work is done, P = W/t

Work is force times distance, W = F*d

Therefore, P = F*d/t = F*v (Speed = distance/time)

a] Power, P = force of winch * v

= mgsin theta * v

= 950*9.8*sin 33 degree *2.05

= 10394.708 Watt

b] The maximum power happens when the force is maximum and the speed is maximum.

  The maximum force (by the winch) happens while the car is acceleratin. When the car is not accelerating, the winch's force is equal to gravity; but in order to make the car accelerate, the winch must pull harder than gravity.

We can use the formula Fnet =ma, to determine how hard the winch is pulling during the acceleration.

Fnet = F winch - F gravity = ma

F winch - mgsin theta = ma

F winch = ma + mgsin theta ( This is the max power provided by winch)

where, a = delta(v)/t = 2.05/12.5 = 0.164 m/s^2

Pmax = (ma+mgsin theta)*v

= (950*0.164 +950*9.8*sin 33 degree)*2.05

= 10714 W

c] Total work, W = mgh

where, h is total length of slope = 1250 sin 33 degree

= 950*9.8*1250*sin 33 degree

= 6.34*10^6 J

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