A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before

releasing her.

The acceleration of gravity is 9.8 m/s

2

.

If he does this 30 times, how much work has

he done?

Answer in units of J.

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times, hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J