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

Answer:

(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) $U=668.16\ J$

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:

$PE=m.g.h$

$PE=24\times 9.8\times 3.3$

$PE=776.16\ J$

Kinetic energy at the bottom of the slide:

$KE=\frac{1}{2} m.v^2$

$KE= 0.5\times 24\times 3^2$

$KE= 108\ J$

(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:

$U=PE-KE$

$U=776.16-108$

$U=668.16\ J$