Asked by Johnathan

Does using (Fcos(theta))d work for this problem or should cos be replaced with sin?

A cheerleader lifts his 66.4 kg partner straight up off the ground a distance of 0.718 m before releasing her.If he does this 27 times, how much work has he done?
Answered by GK
1. The weight of the partner is:
F = mg = (66.4 kg)(9.8 N/kg)
2. The work done in a single lift is:
work = F x distance lifted.
3. Total work = (work /lift)(27 lifts)
Answered by Johnathan
How do you find the force to plug in
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