A cheerleader lifts his 65 kg partner straight up off the ground a distance of 0.628 m before releasing her. Assume the partner's velocity is zero at the beginning and the end of the lift.
The acceleration of gravity is 9.8 m/s?
If he does this 31 times, how much work has he done?
Answer in units of J. Answer in units of J.
To determine the work done by the cheerleader, we need to calculate the gravitational potential energy gained by the partner after being lifted.
The gravitational potential energy can be calculated using the formula:
PE = m * g * h
where m is the mass of the partner, g is the acceleration due to gravity, and h is the height the partner is lifted.
In this case:
m = 65 kg
g = 9.8 m/s^2
h = 0.628 m
PE = 65 kg * 9.8 m/s^2 * 0.628 m
PE = 400.88 J
Since the cheerleader does this 31 times, the total work done is equal to the sum of all the individual works done:
Total work = 31 * PE
Total work = 31 * 400.88 J
Total work = 12,424.08 J
Therefore, the cheerleader has done 12,424.08 J of work.