Question

45.0 m? Assume frictional forces are negligible. (1 hp = 746 W)How long will it take a 2.93 hp motor to lift a 250 kg beam directly upward at constant velocity from the ground to a height of 45.0 m? Assume frictional forces are negligible. (1 hp = 746 W)50.4 s5.26 × 104 s3.76 × 104 s95.8 s

Answer 1

Answer: 50.4 s

Explanation:

The energy expended by the motor, must be equal to the work done by it, which is just equal to the change in the gravitational potential energy of the beam, as we are told that the beam is being lifted directly upward at constant speed (so, the applied force is equal and opposite at any time to the gravity force on the beam) , as follows:

W = m*g*h = 250 kg. * 9.8 m/s²* 45.0 m = 110,250 J

Now, as Power is just Energy / Time, we need just to solve for time this equation.

But, prior to be able to solve this equation, we need that our units be consistent.

As we are given the motor power in HP, we need to convert this unit to Joules/seg, as follows:

1 HP = 746 W  

2.93 HP = 2.93 HP * (746W/1HP) = 2,186 J/s

Solving the Power equation for t:

t = W/P = 110,250 J / 2,186 J/s = 50.4 s