A 68 kgskydiver falls at a constant terminal speed of 59 m/s. (a) At what rate is the gravitational potential energy of the Earth–skydiver system being reduced? (b) At what rate is the system’s mechanical energy being reduced?

The Mass of the sky diver is, m=68 kg

The velocity of the skydiver is, v=59 m/s

The rate of gravitational potential energy is a product of mass, acceleration due to gravity, and velocity. This is used to find the rate of gravitational potential energy.

Formula:

The force of the body using the rate of potential energy, $F\left(x\right)=-\frac{d\cup \left(x\right)}{dxx}$ (1)

Using equation (1), the rate at which gravitational potential energy is reduced is given as:

$$\begin{gathered} F\left(x\right)=\frac{dU\left(x\right)}{{\displaystyle \frac{dt}{{\displaystyle \frac{dx}{dt}}}}} \\ F\left(x\right)=-\frac{dU\left(x\right)}{{\displaystyle \frac{dt}{v_x}}} \\ F\left(x\right)\times v_x=\frac{dU\left(x\right)}{{\displaystyle dt}} \\ m\times g\times v_x=-\frac{dU\left(x\right)}{{\displaystyle dt}} \\ \frac{dU\left(x\right)}{{\displaystyle dt}}=-68kg\times 9.8m/s^2\times 59m/s \\ \frac{dU\left(x\right)}{{\displaystyle dt}}=-3.9\times {10}^{4}J/s \end{gathered}$$

Hence, the value of the rate of energy is $-3.9\times {10}^{4}J/s$.

Since velocity is constant, the rate of change of kinetic energy is zero.

The rate of mechanical energy is the rate of potential energy.

So, the rate at which the mechanical energy is reduced is $-3.9\times {10}^{4}J/s$.