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

`\theta \approx 13.659^(\textdegree)`

Step-by-step explanation:

First, let calculate the kinetic energy of the car:

`K = (1)/(2)\cdot (91\,kg)\cdot (1.29\,(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 - (K)/(m\cdot g \cdot l)`

`\theta = \cos^(-1) \left(1 - (K)/(m\cdot g \cdot l) \right)`

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

`\theta \approx 13.659^(\textdegree)`