Question: You hang by your hands from a tree limb that is a heightabove the ground, with your center of mass a heightabove the ground and your feet a heightabove the ground, as shown in Figure 9.56. You then let yourself fall. You absorb the shock by bending your knees, ending up momentarily at rest in a crouched position with your center of mass a heightabove the ground. Your mass is. You will need to draw labeled physics diagrams for the various stages in the process.

(a) What is the net internal energy change in your body (chemical plus thermal)? (b) What is your speedat the instant your feet first touch the ground? (c) What is the approximate average forceexerted by the ground on your feet when your knees are bending? (d) How much work is done by this force,?

The given data is listed below as follows,

• The limb of the tree is at a distance from the ground is L,
• Initially, the height of the center of mass above the ground is ,h
• The distance of feet above the ground is,d
• Finally, the height of the center of mass above the ground is,b
• The mass of the body of an individual is, M.

The net internal energy of an object is described as the product of the force of that object and the distance moved by the object.

Moreover, internal energy is described as the difference between the heat transfer and the work done by the system.

The equation of the net internal energy is expressed as:

$\Delta E_{int}=F·\left(h-b\right)$ …(i)

Here, F is the force exerted, h is the height of the center of mass above the ground initially and the height of the center of mass above the ground finally.

The equation of the force is expressed as:

$F=M·g$

Here,F is the mass of the body and g is the acceleration due to gravity.

Substitute the above value in equation (i).

$\Delta E_{int}=M·g·\left(h-b\right)$

Thus, the net internal change in energy is as $Mg\left(h-b\right)$.

According to the law of energy conservation, the final and the initial energy are the same.

The equation of the initial and the final energy is expressed as:

$Mgh=Mg\left(h-d\right)+\frac{1}{2}M{v}^2$

Here, is the mass of the body, is the acceleration due to gravity, is the height of the center of mass above the ground initially, is the distance of the feet above the ground and is the instant speed of the body.

The above equation can be reduced as follows:

$$\begin{gathered} gh=g\left(h-d\right)+\frac{1}{2}{v}^2 \\ {v}^2=2gh-2gh+2gd \\ {v}^2=2gd \\ v=\sqrt{2gd} \end{gathered}$$

Thus, the instant speed is when feet touch the ground is$\sqrt{2gd}$$\sqrt{2gd}$.

The equation of the approximate average force is expressed as:

$F=\frac{Mg\left(h-b\right)}{\left(h-d-b\right)}$

Here, is the mass of the body, is the acceleration due to gravity, is the height of the center of mass above the ground initially, is the distance of the feet above the ground and the height of the center of mass above the ground finally.

Thus, the approximate average force is$\frac{Mg\left(h-b\right)}{\left(h-d-b\right)}$ .

The work done by the force is equal to due to the fact that the earth moves very less distance that is negligible, and hence there cannot be any work done.

Thus, the work done by the force exerted by the ground is 0 .