Chapter 9, Problem 39A

Calculate the increase in potential energy when a 20-kg block of ice is lifted a vertical distance of 2 m.

To determine

Increase in potential energy.

Answer to Problem 39A

  $\Delta PE=400\text{J}$

Explanation of Solution

Given:

Mass of ice block, $m=20\text{kg}$

Vertical displacement, $\Delta h=2\text{m}$

Formula Used:

The gravitational potential energy is given as,

  $PE=mgh$

Where, $m$ is mass, $g$ is acceleration due to gravity and $h$ is vertical displacement.

Calculation:

The change in potential energy of the ice block is

  $\Delta PE=mg\Delta h$

Here, $\Delta h$ is change in vertical height.

Take $g=10{\text{m/s}}^2$

On substituting the given values,

  $\begin{array}{l}\Delta PE=20\times 10\times 2\\ =400\text{J}\end{array}$

Conclusion:

Therefore, the change in potential energy is $\Delta PE=400\text{J}$