Question

A 25.0-kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0° from the vertical and releases her from rest. (a) What is her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

Answer 1

(a) 139.7 J

The potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

$U=mg\Delta h$

where

m = 25.0 kg is the mass of the child

g = 9.8 m/s^2

$\Delta h$ is the difference in height between the initial position and the bottom position

We are told that the rope is L = 2.20 m long and inclined at 42.0° from the vertical: therefore, $\Delta h$ is given by

$\Delta h = L - L cos \theta =L(1-cos \theta)=(2.20 m)(1-cos 42.0^{\circ})=0.57 m$

So, her potential energy is

$U=(25.0 kg)(9.8 m/s^2)(0.57 m)=139.7 J$

(b) 3.3 m/s

At the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

$U=K=\frac{1}{2}mv^2$

where

m = 25.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(139.7 J)}{25.0 kg}}=3.3 m/s$

(c) 0

The work done by the tension in the rope is given by:

$W=Td cos \theta$

where

T is the tension

d is the displacement of the child

$\theta$ is the angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. Therefore, $\theta=90^{\circ}$ and $cos \theta=0$, so the work done is zero.