A 19.0 kg child on a 3.00 m long swing is released from rest when the ropes of the the swing make an angle of

Question

A 19.0 kg child on a 3.00 m long swing is released from rest when the ropes of the the swing make an angle of 25.0° with the vertical.

(a) Neglecting friction, find the child's speed at the lowest position.

(b) If the speed of the child at the lowest position is 2.10 m/s, what is the mechanical energy lost due to friction?

Answer 1

1) potential energy = mgh = 19*3.00(1-cos 25)*9.8

At the lowest position all energy is kynetic = 1/2 m v^2

so mgh = 1/2m v^2 and v= sqrt(2gh) = sqrt( 2*9.8*0.2811)=2.35 m/s

Friction losses = 1/2 *19 *(2.35^2-2.10^2)=10.57 joule