"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attached by ropes to a massless pulley on a crane, and four members of the film's construction crew lift the prop at constant speed by delivering 135 W of power each. If 18.0% of the mechanical energy delivered to the pulley is lost to friction, what is the time interval required to lift the spacecraft to the specified height?"

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = 14691.456 J   is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds