Question

Evenin', Physics group. Got a quick question.

The useful power output of a small d.c. motor is used to raise a load of .75 kg through a vertical distance of 1.2 m. The time taken is 18.0 s. The voltage across the motor and the current through it are constant at 6.0 V and .3 A respectively. Assuming that the gravitational force on a mass of 1.0 kg is 10 N, calculate:
i) the power input of the motor (given by the product of the voltage and current of the motor)
ii) the work done
iii) useful power output

Much thanks in advance!

Answer 1

have try to do something????

Answer 2

i) it is already mentioned in question: power input= I*V=.3*6=1.8 watt.
ii) work done = mgh = .75*10*1.2=9 joule
iii) useful power output= work/ time= mgh/t = 9/18= 0.5 watt

Answer 3

Thank you very much! :D