Question

The rope of a swing is 3.00 m long. Calculate the angle from the vertical at which a 91.0 kg man must begin to swing in order to have the same KE at the bottom as a 1530 kg car moving at 1.29 m/s (2.89 mph).

Answer 1

Answer:

$\theta \approx 13.659^{\textdegree}$

Explanation:

First, let calculate the kinetic energy of the car:

$K = \frac{1}{2}\cdot (91\,kg)\cdot (1.29\,\frac{m}{s} )^{2}$

$K = 75.717\,J$

The angle from the vertical is:

$U_{g} = K$

$K = m\cdot g\cdot (1-\cos \theta)\cdot l$

$\cos\theta =1 - \frac{K}{m\cdot g \cdot l}$

$\theta = \cos^{-1} \left(1 - \frac{K}{m\cdot g \cdot l} \right)$

$\theta = \cos^{-1} \left[1 - \frac{75.717\,J}{(91\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (3\,m)} \right]$

$\theta \approx 13.659^{\textdegree}$