# Kelli weighs 420 N, and she is sitting on a playground swing seat that hangs 0.40 m above the...

## Question:

Kelli weighs 420 N, and she is sitting on a playground swing seat that hangs 0.40 m above the ground. Tom pulls the swing back and releases it when the seat is 1.00 m above the ground.

a. How fast is Kelli moving when the swing passes through its lowest position?

b. If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?

## Energy conservation

To find the velocity of the child at the lowest point, we need to take note that the energy of the swing when pulled back and when it is at its lowest point is equal. {eq}PE_i + KE_i = PE_f + KE_f{/eq}

Given

• Weight of Kelli {eq}w = 420 \rm\ N{/eq}
• Height of the of the swing at the lowest position {eq}y_f = 0.4 \rm\ m{/eq}
• Height of the swing when pulled back {eq}y_i = 1 \rm\ m{/eq}

Part A:

Applying the law of conservation of energy to find the velocity of the swing at the lowest point

\begin{align} PE_i + KE_i &= PE_f + KE_f\\ mgy_i &= mgy_f + \frac{1}{2}mv_f^2\\ v_f &= \sqrt{2g(y_i -y_f)}\\ v_f &= \sqrt{2(9.8 \rm\ m/s^2)(1 \rm\ m -0.4 \rm\ m)}\\ v_f &= \boxed{3.43 \rm\ m/s} \end{align}

Part B:

If the velocity is 2.0 meters per second, the work done by friction is

\begin{align} W &= \Delta KE\\ W &= \frac{1}{2}mv^2\\ W &= \frac{1}{2}m(v - v_f)^2\\ W &= \frac{1}{2}(\frac{420 \rm\ N}{9.8 \rm\ m/s^2})(3.43 \rm\ m/s - 2 \rm\ m/s)^2\\ W &= \boxed{43.82 \rm\ J} \end{align}