A 40kg child is swinging on a swing. At the bottom of the swinging motion, the child is 0.6m above the
ground and travelling at a speed of 5m/s. The chains the swing is attached to are 2.1m long.
a) How high above the ground is the child at the highest point of their swinging motion? [2A] [1C]
b) At the highest point of the swinging motion, what is the angle between the chains and their natural
resting position?
To solve this problem, we can apply the principle of conservation of mechanical energy. At the bottom of the swinging motion, the child has gravitational potential energy and kinetic energy. At the highest point of the swinging motion, the child has only gravitational potential energy.
a) Using the principle of conservation of mechanical energy, we can equate the initial mechanical energy to the final mechanical energy.
Initial mechanical energy (at the bottom of the swinging motion) = Final mechanical energy (at the highest point of the swinging motion)
At the bottom of the swinging motion:
Potential energy = mgh
Kinetic energy = (1/2)mv^2
At the highest point of the swinging motion:
Potential energy = mgh'
Since the child is 0.6m above the ground at the bottom of the swinging motion and the chains are 2.1m long, the child is at a height of 0.6m + 2.1m = 2.7m above the ground at the highest point of their swinging motion.
b) At the highest point of the swinging motion, the chains are perpendicular to the ground. This means that the chains are perpendicular to the direction of gravity. Therefore, the angle between the chains and their natural resting position at the highest point of the swinging motion is 90 degrees.