Question

A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What energy transfers and transformation occurs during the slide?b.What is the total change in the thermal energy of the slide and the seat of her pants?

Answer 1

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b)
`U=668.16\ J`

Step-by-step explanation:

Given:

• mass of the child,
  `m=24\ kg`

• height of the slide,
  `h=3.3\ m`

• initial velocity of the child at the slide,
  `v_i=0 m.s^(-1)`

• final velocity of the child at the bottom of slide,
  `v_f=3\ m.s^(-1)`

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

`PE=m.g.h`

`PE=24* 9.8* 3.3`

`PE=776.16\ J`

Kinetic energy at the bottom of the slide:

`KE=(1)/(2) m.v^2`

`KE= 0.5* 24* 3^2`

`KE= 108\ J`

(b)

Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.

This can be given as:

`U=PE-KE`

`U=776.16-108`

`U=668.16\ J`