How much work?

Question

A dancer lifts his 53.6 kg partner straight
up off the ground a distance of 0.373 m before
releasing her.
The acceleration of gravity is 9.8 m/s2

If he does this 22 times, how much work has
he done? Answer in units of J.

thank you in advance for helping ^_^

Answer 1

1...Force = Mass x Acceleration.
= 53.6kg x 9.81m/s²
= 53.6 x 9.81 = 525.8 N (Force).
2...Work = Force x Distance
= 525.8 N x 0.373m = 196.13 J.
196.13 x 22 = 4,315J. (If he drops her to her feet).
(However, if he lifts AND lowers her, it will be double this work..8,630J).

Answer 2

The only work he has done is WE = mgh per lift and per return; where m = 53.6 kg, h = .373 meters, and g = 9.8 m/sec^2. Here's why.

To lift the dancer, he does total energy TE = PE = mgh work because he is overcoming the force of gravity W = mg to create potential energy PE through that work. So he works WL = 22 mgh in lifting his partner 22 times.

To return the dancer, he does TE = 0 = PE - WE; so we have PE = WE = mgh work to ensure she lands lightly (with little or no velocity) upon impact with the dance floor. We assume TE = 0 because the potential energy PE is reduced to zero from the work WE done in returning the dancer softly to the floor. Thus WR = 22 mgh in returning his partner to the floor 22 times.

Therefore, the total work done is W = WR + WL = 44 mgh; you can do the math.

This is a conservation of energy problem. The key is that when the partner is at h, the total energy in the partner is potential energy PE = mgh. And that PE came about by the work the lifter put into lifting his partner.

Also key, the total energy on the floor TE = 0 because there is no potential energy and no kinetic energy. And the only way PE could be zero (when it started out as PE = mgh) is to do work WE returning his partner to the floor, equal to the starting PE.

PS: Thought I'd add this for a bit more accuracy. When putting the partner back to the floor, there is in fact some kinetic energy. That's clear because to go from h to 0 height, there is movement, which means there is some velocity and, therefore, KE > 0. But the velocity is so slow, the soft landing part, that KE is quite small. In other words, some of the work putting the partner down is converted into a little bit of ke, but mostly the work converts the PE at h to zero PE on the floor.

Answer 3

Work done = Force X distance moved in direction of force .

work done = (53.6X9.8)X(0.373X22)
Wrk done = 4262.1964 J