Chapter 8, Problem 1P

To determine

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

Answer: 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

Explanation:

Given information:

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 $8.2$ is,

$\Delta K+\Delta U=0$

• $\Delta K$ is the change in kinetic energy of the ball.
• $\Delta U$ 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.

Formula to calculate the velocity of the ball just before strike the earth by energy of conservation is,

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

• $K_1$ 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.
• $U_2$ is the final potential energy of the ball.

Formula to calculate final kinetic energy of the ball is,

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

• $m$ is the mass of the ball.

$v_{\text{f}}$ is the final velocity of the ball.

Formula to calculate initial potential energy is,

$U_1=mgh$

$g$ is the acceleration due to gravity.

$h$ 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 $\sqrt{2gh}$.