Chapter 11, Problem 7STP

To determine

To Choose: The correct option.

Answer to Problem 7STP

Option C

Explanation of Solution

Given:

Mass of the ball, $m=6.0\times {10}^{-2}\text{kg}$

Height from which the ball is dropped, $h=1.0\text{m}$

Energy decreases by, $\Delta E=0.14\text{J}$

Formula Used:

Mechanical energy is conserved. So, change in kinetic energy (at the bottom) =change in gravitational potential energy.

  $\frac{1}{2}m{v}^2=mgh$

Where, m is the mass, g is the gravitational acceleration, h is the height, u is the initial velocity and v is the final velocity

Calculations:

The kinetic energy decreases by $\Delta E=0.14\text{J}$ after it bounces from the flat surface.

  $\begin{array}{c}{E}_2=\frac{1}{2}m{v}^2-0.14\text{J}\\ =mgh-0.14\text{J}\end{array}$

Substitute the values and solve:

  $\begin{array}{c}{E}_2=mgh-0.14\text{J}\\ \text{=}\left(6.0\times {10}^{-2}\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(1.0\text{m}\right)-0.14\text{J}\\ \text{= 0}\text{.588}\text{J-0}\text{.14 J}\\ \text{=}\text{0}\text{.448 J}\\ \approx \text{ 0}\text{.45 J}\end{array}$

Conclusion:

Thus, correct option is C.