Question

Tarzan, whose mass is 100kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.9m above the ground and the bottom of his dangling feet are at a height 2.1 above the ground. When he first hits the ground he has dropped a distance 2.1, so his center of mass is (2.9-2.1) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height above the ground. (a) Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground? (b) Consider the extended system. What is the net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position?

Short answer

(a) Thespeed at the instant just before Tarzan's feet touch the ground is 6.42m/s .

(b) The net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position is -2354.82J .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of Tarzan is,$m=100\mathrm{kg}$ .
• The initial speed of Tarzan at rest is,u=0
• The height of dangling feet above the ground is,$h_1=2.1\mathrm{m}$ .
• The height of the center of mass above ground is,$h=2.9\mathrm{m}$ .
• The height of the center of mass on the ground is,$h_{2}2.9-2.1=0.8\mathrm{m}$ .

The crouched position of the above-ground is,$h_3=0.5\mathrm{m}$ .

Significance of equation of motion

The equation of motion is defined as the simple formula that calculates the displacement, velocity (initial or final), and acceleration of the object relative to the frame of reference.

(a) Evaluation of the speed

Using $3^{rd}$equation of motion is given as follows,

$v^2=u^2+2g{h}_1$

Here u is the initial speed, v is the final speed, and g is the acceleration due to gravity.

Substitute 0 for u,$9.8\mathrm{m}/{\mathrm{s}}^2$ for g, and 2.1m forrole="math" localid="1656908484636" $h_1$ in the above equation.

$$\begin{gathered} v^2=0+2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2.1\mathrm{m}\right) \\ \mathrm{v}=\sqrt{41.16} \\ \mathrm{v}=6.42\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed at the instant just before Tarzan's feet touch the ground is 6.42m/s .

(b) Determination of the change in the internal energy

The initial internal energy is given as,

$$\begin{gathered} U_1=PE+KE \\ =mg{h}_2+\frac{1}{2}m{v}^2 \end{gathered}$$

Substitute 100kg for m,$9.8m/s^2$ for g,0.8m for$h_2$ and 6.42m/s for v in the above equation.

$$\begin{gathered} U_1=\left[\left(100\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.8\mathrm{m}\right)+\frac{1}{2}\left(100\mathrm{kg}\right){\left(6.42\mathrm{m}/\mathrm{s}\right)}^2\right]\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =2844.82\mathrm{J} \end{gathered}$$

Now, the final internal energy is given as,

$U_2=mg{h}_3$

Substitute 100 kg for m,$9.8m/s^2$ for g, and 0.5m for$h_3$ in the above equation.

$$\begin{gathered} U_2=\left(100\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.5\mathrm{m}\right)\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =490\mathrm{J} \end{gathered}$$

The net change in the internal energy is,

$∆U=∆U_2-U_1$

Substitute 490J for $U_2$, and 2844.82 J for$U_1$ in the above equation.

$$\begin{gathered} ∆U=490\mathrm{J}-2844.82\mathrm{J} \\ =-2354.82\mathrm{J} \end{gathered}$$

Thus, the net change in internal energy for Tarzan from just before his feet touch the ground to when he is in the crouched position is$-2354.82\mathrm{J}$.