Gravitational Potential Energy Near a Surface (College Board AP® Physics 1: Algebra-Based): Revision Note

A rock of mass $2.1\mathrm{kg}$ is dropped from a cliff into the water $12\mathrm{m}$ below. The person that drops the rock releases it from a height of $1.4\mathrm{m}$ above the ground. Which of the following values is correct for the speed of the rock just before it hits the water?

A: $6.5\mathrm{m}/\mathrm{s}$

B: $14.5\mathrm{m}/\mathrm{s}$

C: $15.5\mathrm{m}/\mathrm{s}$

D: $16.4\mathrm{m}/\mathrm{s}$

The correct answer is D

Step 1: Analyze the scenario

• The cliff is $12\mathrm{m}$ above the water, but the rock is dropped from a height of $1.4\mathrm{m}$ above the ground; therefore, the total vertical distance the rock travels is $12+1.4=13.4\mathrm{m}$

• Air resistance is always assumed to be negligible unless otherwise stated

• Therefore, it can be assumed that the gravitational potential energy is transformed into kinetic energy with 100% efficiency

Step 2: List the known quantities

• Mass of rock, $m=2.1\mathrm{kg}$

• Vertical distance traveled by rock, $∆y=13.4\mathrm{m}$

• Magnitude of the gravitational field strength at the Earth's surface, $g=10\mathrm{N}/\mathrm{kg}$

Step 3: Calculate the change in gravitational potential energy as the rock falls to the surface of the water

$∆U_g=mg∆y$

$∆U_g=2.1·10·13.4=281.4\mathrm{J}$

Step 4: Equate the change in gravitational potential energy to the gain in kinetic energy

$∆U_g=K=\frac{1}{2}m{v}^2$

Step 5: Rearrange to make speed the subject and substitute in the known values to calculate

$∆U_g=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{2·∆U_g}{m}}$

$v=\sqrt{\frac{2·281.4}{2.1}}=16.4\mathrm{m}/\mathrm{s}$

You could also solve this problem by equating the expressions for gravitational potential energy and translational kinetic energy:

$mg∆y=\frac{1}{2}m{v}^2$

$g∆y=\frac{1}{2}{v}^2$

$v=\sqrt{2g∆y}$