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 \(-g\). \[y(t)=h-\frac{1}{2} g t^{2} \notag\] 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 \(d K / d t=m g v\), while the potential energy \(U\) decreases at the rate \(d U / d t=-m g v\). Since \(v\) increases as the body...So as the body falls, its kinetic energy \(K\) increases at the rate \(d K / d t=m g v\), while the potential energy \(U\) decreases at the rate \(d U / d t=-m g v\). 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, \(d K / d t=-d U / d t=0\); after the body falls through a height \(h\), the rates of change have increased to \(d K / d t=-d U / d t=m \sqrt{2 g^{3} h}\).
