Chapter 11, Problem 65A

(a)

To determine

The potential energy of the rock-Earth system relative to the base of the cliff.

Answer to Problem 65A

The potential energy of the rock-Earth system relative to the base of the cliff is $2\times {10}^4\text{ J}$ .

Explanation of Solution

Given:

The mass of the rock is $m=20\text{ kg}$

The height of the cliff is $h=100\text{ m}$ .

Formula used:

The expression for the potential energy is,

$PE=mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the rise.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

The potential energy of the rock-Earth system relative to the base of the cliff is

$\begin{array}{c}PE=mgh\\ PE=\left(20\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(\text{100 m}\right)\\ =2\times {10}^4\text{ J}\end{array}$

Conclusion:

Thus, the potential energy of the rock-Earth system relative to the base of the cliff is $2\times {10}^4\text{ J}$ .

(b)

To determine

The kinetic energy just before it strikes the ground.

Answer to Problem 65A

The kinetic energy just before it strikes the ground is $2.55\times {10}^8\text{ J}$ .

Explanation of Solution

Given:

The mass of the rock is $m=20\text{ kg}$

The height of the cliff is $h=100\text{ m}$ .

Calculation:

From part (a), the potential energy of the rock-Earth system relative to the base of the cliff is $P{E}_{\text{f}}=2\times {10}^4\text{ J}$ .

Consider the initial potential energy for the system $P{E}_{\text{i}}=0$ .

The kinetic energy just before it strikes the ground is,

$\begin{array}{l}KE=\Delta PE\\ KE=P{E}_{\text{f}}-P{E}_{\text{i}}\\ KE=2\times {10}^4\text{ J}-0.0\text{ J}\\ KE=2\times {10}^4\text{ J}\end{array}$

Conclusion:

Thus, the kinetic energy just before it strikes the ground is $2\times {10}^4\text{ J}$ .

(c)

To determine

The speed of the rock as it hits the ground.

Answer to Problem 65A

The speed of the rock as it hits the ground is $45\text{m}/\text{s}$ .

Explanation of Solution

Given:

The mass of the rock is $m=20\text{ kg}$

The height of the cliff is $h=100\text{ m}$ .

Formula used:

The expression for the kinetic energy as follows:

$KE=\frac{1}{2}m{v}^2$

Here, $m$ is the mass and $v$ is the speed.

Calculation:

From part (b), the kinetic energy just before it strikes the ground is $KE=2\times {10}^4\text{ J}$ .

The speed of the rock is,

$\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ v=\sqrt{\frac{2KE}{m}}\\ v=\sqrt{\frac{2\times \left(2\times {10}^4\text{ J}\right)}{\left(20\text{ kg}\right)}}\\ v=45\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the speed of the rock as it hits the ground is $45\text{m}/\text{s}$ .