Question

A 200. Kg load of bricks is to be lifted from the ground to the top of a 30.0 m high building in 30.0 s. If a 75.0 N frictional force in the pulley opposes motion, what minimum power motor would be needed?

Answer 1

The minimum power the motor would need is 2035 Watts.

Step-by-step explanation:

the minimum power motor needed is given by the total work done on the load, Wtot and the total time it takes the load to be lifted, given by:

P = Wtot/t

the forces doing works on the load is the gravitational force, Wfg and the frictional force, Wf the total work done on the load is:

Wtot = Wfg + Wf

= m×g×d + f×d

= d×(m×g + f)

= (30)×(200×9.8 + 75)

= 61050 J

it takes t = 30s for the load to be lifted high the building, the minimum power neeed is then:

P = Wtot/t

= (61050)/(30)

= 2035 Watts

Therefore, the minimum power the motor would need is 2035 Watts.