Gravitational Potential Energy Change in Free Fall

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Work and Energy Beginner mechanical energy and conservation

Source: High school physics (Chinese)

Problem Sets:

work/energy basic work-energy 1018

Problem

An object with a mass of $m = 1$ kg is in free fall. Take the acceleration due to gravity as $g = 10$ m/s².

  1. What is the decrease in its gravitational potential energy during the first second of fall?
  2. What is the decrease in its gravitational potential energy during the second second of fall?

\Delta E_{p1} = 50 J \Delta E_{p2} = 150 J

The decrease in gravitational potential energy is equal to the work done by gravity, $\Delta E_p = W_g = mg\Delta h$. The distance fallen from rest is given by $\Delta h = \frac{1}{2}gt^2$.

For the first second (from $t=0$ s to $t=1$ s):

$$\Delta h_1 = \frac{1}{2}g(1)^2 - \frac{1}{2}g(0)^2 = \frac{1}{2}g$$ $$\Delta E_{p1} = mg\Delta h_1 = \frac{1}{2}mg^2$$ $$\Delta E_{p1} = \frac{1}{2}(1 \text{ kg})(10 \text{ m/s}^2)^2 = 50 \text{ J}$$

For the second second (from $t=1$ s to $t=2$ s):

$$\Delta h_2 = \frac{1}{2}g(2)^2 - \frac{1}{2}g(1)^2 = \frac{3}{2}g$$ $$\Delta E_{p2} = mg\Delta h_2 = \frac{3}{2}mg^2$$ $$\Delta E_{p2} = \frac{3}{2}(1 \text{ kg})(10 \text{ m/s}^2)^2 = 150 \text{ J}$$