Question

A 36.0 kg child starting from rest slides down a water slide with a vertical height of 14.5 m. (Neglect friction.)(a)What is the child's speed halfway down the slide's vertical distance? answer in:____m/s(b)What is the child's speed three-fourths of the way down? answer in:____m/s

Answer 1

Answer:

a) The child's speed halfway down the slide's vertical distance = 11.92 m/s

b) The child's speed three-fourths of the way down = 8.43 m/s

Step-by-step explanation:

The mass, m = 36 kg

The vertical height, h = 14.5 m

a) At halfway down the height, the kinetic energy and the potential energy are equal

`\begin{gathered} (1)/(2)mv^2=mg((h)/(2)) \\ \\ 0.5mv^2=m(g)((14.5)/(2)) \\ \\ 0.5v^2=(9.8)(7.25) \\ \\ 0.5v^2=71.05 \\ \\ v^2=(71.05)/(0.5) \\ \\ v^2=142.1 \\ \\ v=√(142.1) \\ \\ v=11.92\text{ m/s} \end{gathered}`

b) To get the child's speed three-fourths of the way down

`\begin{gathered} mg(h-(3h)/(4))=(1)/(2)mv^2 \\ \\ mg((h)/(4))=m(v^2)/(2) \\ \\ v^2=g(h)/(2) \\ \\ v^2=(9.8(14.5))/(2) \\ \\ v^2=71.05 \\ \\ v=√(71.05) \\ \\ v=8.43\text{ m/s} \end{gathered}`

The child's speed three-fourths of the way down = 8.43 m/s