Question

A 34.9-kg child starting from rest slides down a water slide with a vertical height of 16.0 m. (Neglect friction.)

Answer 1

Answer:

a) 12.528 m/s

b) 15.344 m/s

Explanation:

Given:

Mass of the child, m = 34.9 kg

Height of the water slide, h = 16.0 m

Now,

a) By the conservation of energy,

loss in potential energy = gain in kinetic energy

mgh = $\frac{\textup{1}}{\textup{2}}\textup{m}\times\textup{v}^2$

where,

g is the acceleration due to the gravity

v is the velocity of the child

thus,

at halfway down, h = $\frac{\textup{16}}{\textup{2}}$= 8 m

therefore,

34.9 × 9.81 × 8 = $\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2$

or

v = 12.528 m/s

b)

at three-fourth way down

height = $\frac{\textup{3}}{\textup{4}}\times16$ = 12 m

thus,

loss in potential energy = gain in kinetic energy

34.9 × 9.81 × 12 = $\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2$

or

v = 15.344 m/s

Answer 2

Answer:

(a) 12.52 m/s

(b) 15.34 m/s

Explanation:

mass, m = 34.9 kg

h = 16 m

(a) Initial velocity, u = 0

height = h / 2 = 16 / 2 = - 8 m (downward)

let the speed of child is v.

acceleration, a = - 9.8 m/s^2 (downward)

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2}=0^{2}+2\times 9.8\times 8$

v = 12.52 m/s

(b) Initial velocity, u = 0

height = 3 h / 4 = 12 m = - 12 m (downward)

let the speed of child is v.

acceleration, a = - 9.8 m/s^2 (downward)

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2}=0^{2}+2\times 9.8\times 12$

v = 15.34 m/s