A cheerleader lifts his 43.8 kg partner straight up off the ground a distance of 0.737 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 2. If he does this 27 times, how much work has he done? Answer in units of J.

The cheerleader does a total work of 8,532 J (8,541 J without any intermediate rounding)

Explanation:

Hi there!

The equation of work is the following:

W = F · d

Where:

W = work.

F = applied force.

d = distance.

The cheerleader's partner is on the ground due to the force of gravity acting on her, also known as weight. To lift the partner at a constant velocity, the cheerleader has to suppress this force applying on the partner a force that is equal to the partner's weight but in opposite direction.

The weight of the partner is calculated as follows:

Weight = m · g

Where m is the mass of the partner and g is the acceleration due to gravity.

Let's calculate the weight of the partner:

Weight = 43.8 kg · 9.8 m/s²

Weight = 429 N

]Then, the work done in one lift by the cheerleader is the following:

W = F · d

W = 429 N · 0.737 = 316 J

If the cheerleader does 27 lifts, the work done will be (316 J · 27) 8,532 J