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)
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?
2 answers
How do you find the force to plug in
1 answer