Given a body of mass m released from rest at time t=0 from a height h above the floor, we find the following results about the motion. Here the y axis points upward and has its origin ...Given a body of mass m released from rest at time t=0 from a height h above the floor, we find the following results about the motion. Here the y axis points upward and has its origin at the floor, and so the acceleration due to gravity is −g. y(t)=h−12gt2 U(t) & =m g h-\frac{1}{2} m g^{2} t^{2} \\ \frac{d K}{d t} & =-m g v=m g^{2} t=m g \sqrt{2 g(h-y)} \\ \frac{d U}{d t} & =m g v=-m g^{2} t=-m g \sqrt{2 g(h-y)} \\
So as the body falls, its kinetic energy K increases at the rate dK/dt=mgv, while the potential energy U decreases at the rate dU/dt=−mgv. Since v increases as the body...So as the body falls, its kinetic energy K increases at the rate dK/dt=mgv, while the potential energy U decreases at the rate dU/dt=−mgv. Since v increases as the body falls, the rate of change of the kinetic and potential energies increases as the body falls. Right after the body is released, dK/dt=−dU/dt=0; after the body falls through a height h, the rates of change have increased to dK/dt=−dU/dt=m√2g3h.
