Question

The rope of a swing is 3.10 m long. Calculate the angle from the vertical at which a 76.0 kg man must begin to swing in order to have the same KE at the bottom as a 1520 kg car moving at 1.01 m/s (2.26 mph).

Answer 1

Answer:

The angle from the vertical is 48.72°

Explanation:

Given :

Length of rope $l = 3.10$ m

Mass of man $m = 76$ Kg

Mass of car $M = 1520$ Kg

Velocity of car $v = 1.01$ $\frac{m}{s}$

According to conservation law,

Potential energy of man is converted to kinetic energy of car moving,

   $\frac{1}{2} M v^{2} = mgh$

We calculate height,

   $h = \frac{M v^{2} }{2mg}$

   $h = \frac{1520(1.01)^{2} }{2\times 75 \times 9.8}$                      ( ∵ $g = 9.8 \frac{m}{s^{2} }$ )

   $h = 1.05$ m

This is the distance of the rope at the bottom,

So we take difference of it.

⇒ $3.10 - 1.05 = 2.05$

We can calculate angle between them,

$\cos \theta = \frac{2.05}{3.10} = 0.6597$

 $\theta =$ 48.72°

Therefore, the angle from the vertical is 48.72°