Question

Tarzan swings on a 29.0-m-long vine initially inclined at an angle of 34.0° with the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 4.00 m/s?

Answer 1

The energy conservation law:

$$T + U = \text{const}$$ $$T = \frac{mv^2}{2} - \text{kinetic energy}$$

m - mass of the body

v - speed

$$U = mgh - \text{potential energy}$$

g - gravitational acceleration

h - high

$$T_1 + U_1 = T_2 + U_2$$

1 - initial state

2 - final state

a.

$$T_1 = 0 \quad U_1 = -mgl * \cos(34)$$ $$U_2 = -mhl$$ $$T_2 = T_1 + U_1 - U_2 = mgl(1 - \cos(34))$$ $$\frac{mv^2}{2} = mgl(1 - \cos(34)) \quad \Rightarrow \quad v = \sqrt{2gl(1 - \cos(34))}$$ $$v = 9.8 \, \text{m/s}$$

b.

$$T_1 = \frac{mv_0^2}{2}$$ $$T_2 = T_1 + U_1 - U_2 = \frac{mv_0^2}{2} + mgl(1 - \cos(34))$$ $$\frac{mv^2}{2} = \frac{mv_0^2}{2} + mgl(1 - \cos(34))$$ $$v = \sqrt{v_0^2 + 2gl(1 - \cos(34))}$$ $$v = 10.6 \, \frac{\text{m}}{\text{s}}$$