# A 65 kg child sits in a swing suspended with 2.6 m long ropes. The swing is held aside so that...

## Question:

A {eq}65\ kg {/eq} child sits in a swing suspended with {eq}2.6\ m {/eq} long ropes. The swing is held aside so that the ropes make an angle of {eq}34 ^\circ {/eq} with the vertical. Use conservation of energy to determine the speed the child will have at the bottom of the arc when she is let go.

## Law of conservation of energy

The energy in an isolated system remained unchanged or it is constant. Mechanical energy (sum of kinetic and potential energy) is conserved in this problem. Initially, the child has no kinetic energy and after the swing was let go, the child now has kinetic energy

Given:

• Mass {eq}m = 65 \rm\ kg{/eq}
• Length of the ropes {eq}l = 2.6 \rm\ m{/eq}
• Angle {eq}\theta = 34 ^{\circ}{/eq}

The height in which the swing is raise

\begin{align} y_i &= l - lcos\theta\\ y_i &= 2.6 \rm\ m - (2.6 \rm\ m)cos (34 ^{\circ})\\ y_i &= 0.44 \rm\ m \end{align}

Applying the conservation of energy to find the velocity at the bottom

\begin{align} \rm\ E_i &= E_f\\ mgy_i + \frac{1}{2}mv_i^2 &= mgy_f + \frac{1}{2}mv_f^2\\ v_f &= \sqrt{2gy_i}\\ v_f &= \sqrt{2 (9.8 \rm\ m/s^2) (0.44 \rm\ m)}\\ v_f &= \boxed{2.94 \rm\ m/s} \end{align}