Question

A playground slide is 8.80 ft long and makes an angle of 25.0° with the horizontal. A 63.0-kg child, initially at the top, slides all the way down to the bottom of the slide. (a) Choosing the bottom of the slide as the reference configuration, what is the system's potential energy when the child is at the top and at the bottom of the slide? What is the change in potential energy as the child slides from the top to the bottom of the slide? (Include the sign of the value in your answer.)

Answer 1

initial: 1654.6 J, final: 0 J, change: -1654.6 J

Step-by-step explanation:

The length of the slide is

L = 8.80 ft = 2.68 m

So the height of the child when he is at the top of the slide is (with respect to the ground)

`h = L sin \theta = (2.68 m)sin 25.0^(\circ)=1.13 m`

The potential energy of the child at the top is given by:

`U = mgh`

where

m = 63.0 kg is the mass of the child

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.13 m

Substituting,

`U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J`

At the bottom instead, the height is zero:

h = 0

So the potential energy is also zero: U = 0 J.

This means that the change in potential energy as the child slides down is

`\Delta U = 0 J - (1654.6 J) = -1654.6 J`