Question

1. a 1.00 kg pendulum bob is released from a height of 0.200 m. its speed at the bottom of its swing is 1.95 m/s on the first pass. how much energy is lost due to friction during one complete swing of the pendulum?

Explanation:

Step1: Calculate initial potential energy

The initial potential energy $U = mgh$, where $m = 1.00\ kg$, $g=9.8\ m/s^{2}$, $h = 0.200\ m$. So $U=1.00\times9.8\times0.200 = 1.96\ J$.

Step2: Calculate kinetic energy at the bottom - first pass

The kinetic energy $K=\frac{1}{2}mv^{2}$, with $m = 1.00\ kg$ and $v = 1.95\ m/s$. Then $K=\frac{1}{2}\times1.00\times(1.95)^{2}=\frac{1}{2}\times1.00\times3.8025 = 1.90125\ J$.

Step3: Calculate energy lost

The energy lost $\Delta E=U - K$. So $\Delta E=1.96 - 1.90125=0.05875\ J$.

Answer:

$0.05875\ J$