Chapter 8, Problem 1P

(a)

To determine

The appropriate version of the equation $8.2$ for the ball and the earth system and the speed of the ball earth just before it strike the earth.

Answer to Problem 1P

The appropriate version of equation $8.2$ for the ball and the earth is $\Delta K+\Delta U=0$ and the speed of the ball just before it strike the earth is $\sqrt{2gh}$.

Explanation of Solution

A ball of mass $m$ falls from a height $h$ to the floor.

From the conservation of energy, the total energy of an isolated system remains conserved. It means if kinetic energy increases then the potential energy decreases.

The appropriate version of the equation $8.2$ for the ball and the earth

  $\Delta K+\Delta U=0$

Here, $\Delta K$ is the change in kinetic energy of the ball and $\Delta U$ is the change in potential energy.

As the system is under free fall therefore initial velocity of the ball is zero and its final potential energy will also be zero.

Write the formula to calculate the velocity of the ball just before strike the earth by energy of conservation

    $\begin{array}{c}\Delta K+\Delta U=0\\ \left(K_2-K_1\right)+\left(U_2-U_1\right)=0\end{array}$        (I)

Here, $K_1$ is the initial kinetic energy of the ball, $K_2$ is the final kinetic energy of the ball, $U_1$ is the initial potential energy of the ball and $U_2$ is the final potential energy of the ball.

Write the formula to calculate final kinetic energy of the ball

    $K_2=\frac{1}{2}m{v}_{\text{f}}{}^2$

Here, $m$ is the mass of the ball and $v_{\text{f}}$ is the final velocity of the ball.

The initial kinetic energy of the ball is zero.

Write the formula to calculate initial potential energy

    $U_1=mgh$

Here, $g$ is the acceleration due to gravity and $h$ is the distance between ball and floor.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_2$, $mgh$ for $U_1$, $0$ for $K_1$ and $0$ for $U_2$ in equation (I).

    $\begin{array}{c}\left(\frac{1}{2}m{v}_{\text{f}}{}^2-0\right)+\left(0-mgh\right)=0\\ v_{\text{f}}{}^2=2gh\\ v_{\text{f}}=\sqrt{2gh}\end{array}$

Conclusion:

Therefore, the appropriate equation $8.2$ is $\Delta K+\Delta U=0$ and the speed of the ball just before it strike the earth is $\sqrt{2gh}$.

(b)

To determine

The appropriate version of the equation $8.2$ for the system and the speed of the ball just before it strike the earth.

Answer to Problem 1P

The appropriate version of equation $8.2$ is $\Delta K+\Delta U=0$ and the speed of the ball just before it strike the earth is $\sqrt{2gh}$.

Explanation of Solution

A ball of mass $m$ falls from a height $h$ to the floor.

From the conservation of energy, the total energy of an isolated system remains conserved. It means if kinetic energy increases then the potential energy decreases.

The appropriate version of the equation $8.2$ for the ball is,

  $\Delta K=W$

Here, $\Delta K$ is the change in kinetic energy of the ball and $W$ is the work done by the ball.

As the system is under free fall therefore initial velocity of the ball is zero and its final potential energy will also be zero.

Rewrite the above equation

  $K_2-K_1=W$

Write the formula to calculate initial potential energy

    $W=mgh$

Here, $g$ is the acceleration due to gravity and $h$ is the distance between ball and floor.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_2$ and $mgh$ for $W$, $0$ for $K_1$ and $0$ for $U_2$ in equation (I).

    $\begin{array}{c}\left(\frac{1}{2}m{v}_{\text{f}}{}^2-0\right)=mgh\\ v_{\text{f}}{}^2=2gh\\ v_{\text{f}}=\sqrt{2gh}\end{array}$

Conclusion:

Therefore, the appropriate equation $8.2$ is $\Delta K+\Delta U=0$ and the speed of the ball just before it strike the earth is $\sqrt{2gh}$.