Question

A 24.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 41.0 ∘ from the vertical and releases her from rest.

What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?

How fast will she be moving at the bottom of the swing?

How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

Answer 1

Potential energy of the child just as she is released, is more than the potential energy at the bottom by ' mgh'
where h is the height gained, compared to bottom most point.
h = L( 1 - cos (theta) ) = 0.515 m
hence potential energy gained is 24x9.8x0.515 = 121.15J
At bottom this potental energy converts into kinetic energy, hence speed at the bottom is given by
1/2 mv² = 121.15

so v = 3.82 m/s

Work done by tension is zero, as force of tension at any instant is normal to the velocity of child, and power of force is given by F.V = FV cos (theta).
Angle theta at any instant is zero, so is the power or work of the tension