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marshall27 [118]
3 years ago
5

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).
Physics
1 answer:
daser333 [38]3 years ago
3 0

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}

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