A 3.2 kgsloth hangs 3.0 mabove the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point y=0to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

The mass of the sloth hanging is,$m=3.2\mathrm{kg}$

The height of the sloth hanging from the ground is $h=3.0\mathrm{m}$.

By using the equation for gravitational potential U, we can find thegravitational potential energy U. Here, mechanical energy is conserved. So, kinetic energy Kis equal to the gravitational potential energy U. Finally, by using the equation for kinetic energy K, we can find thespeedvof the sloth just before it reachesthe ground.

Formula:

The gravitational potential is,$U=mgh$

The kinetic energy is,$K=\frac{1}{2}m{v}^2$

$$\begin{gathered} U=mgh \\ =3.2\times 9.8\times 3.0 \\ =94\mathrm{J} \end{gathered}$$

Hence, the gravitational potential energy is, U =94 J

The mechanical energy is conserved; hence, we get the kinetic energyKis

$$\begin{gathered} K=U \\ K=94\mathrm{J} \end{gathered}$$

Hence, the kinetic energy is,K=94 J

We know that kinetic energy is given by

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ {v}^2=\frac{2K}{m} \end{gathered}$$

Therefore, we get the speed v as

$$\begin{gathered} v=\sqrt{\frac{2K}{m}} \\ =\sqrt{\frac{2\times 94}{3.2}} \\ =7.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the sloth just before it reaches the ground is,v=7.7 m/s