Question

The mass of a child is 35kg. The child is raised to point A which is 1.5m and then released. She swings downwards through the equilibrium position B at 0.80m. What is the loss in gravitational potential energy of the child between A and B. And assuming that air resistance is neglible, calculate the speed of the child as she passes through the equilibrium position B

Answer 1

change in potential energy = mgh where h=(1.5-0.80) =0.70 meter
this energy will transfer to the kinetic energy of the child at point B.
so..mgh=1/2mv^2 v= the velocity at point B