Question

A 20 kg child is on a swing that hangs from 3.0-m-long chains. What is her maximum speed if she swings out to a 45° angle?

Short answer

As a result, the child's maximum speed is$4.15\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Introduction

The distance travelled by an object per unit time is known as its speed.

Explanation

Draw the child's and swing's free body diagrams.

Calculate the child's increased height.

Putting $3\mathrm{m}$for ll, $0.7071$for $\cos 45°$in equation $x=l\cos 45°$

$x=(3\mathrm{m})(0.7071)$

$=2.12\mathrm{m}$

Putting $2.12\mathrm{m}$for $x$, and $3\mathrm{m}$for $l$in equation $h=l-x$

$h=3\mathrm{m}-2.12\mathrm{m}$

$=0.88\mathrm{m}$

Calculate the child's maximum speed.

At the two positions depicted in the diagram, apply the law of conservation of energy.

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$

$\frac{1}{2}m{v}_{\max }^2+0=0+mgh$

$v_{\max }=\sqrt{2gh}$

Putting $9.81\mathrm{m}/{\mathrm{s}}^2$for $g$, and $0.88\mathrm{m}$for $h$by equation.

$v_{\max }=\sqrt{2\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)(0.88\mathrm{m})}$

$=\sqrt{17.26{\mathrm{m}}^2/{\mathrm{s}}^2}$

$=4.15\mathrm{m}/\mathrm{s}$

As a result, the child's maximum speed is $4.15\mathrm{m}/\mathrm{s}$.