Physics: Energy, Climate, and Sustainability - Mechanical Energy Calculations and Conservation of Energy

A $2 \text{ kg}$ book is lifted from the floor to a shelf $1.5 \text{ m}$ high. Calculate the Gravitational Potential Energy ($E_p$) gained by the book. (Take $g = 9.8 \text{ m/s}^2$)

$E_p = mgh$ $E_p = 2 \times 9.8 \times 1.5$ $E_p = 29.4 \text{ J}$

The gain in potential energy is calculated by multiplying the mass, the gravitational constant, and the vertical height increased.

A ball with a mass of $0.5 \text{ kg}$ is rolling at a velocity of $4 \text{ m/s}$. Find its Kinetic Energy ($E_k$).

$E_k = \frac{1}{2}mv^2$ $E_k = \frac{1}{2} \times 0.5 \times (4)^2$ $E_k = 0.25 \times 16$ $E_k = 4 \text{ J}$

Kinetic energy is determined by the mass and the square of the velocity. Even a small increase in speed significantly increases the kinetic energy.

An object of mass $5 \text{ kg}$ is dropped from a height of $20 \text{ m}$. Using the Law of Conservation of Energy, calculate its velocity just before it hits the ground. (Ignore air resistance, $g = 10 \text{ m/s}^2$)

Initial $E_p = mgh = 5 \times 10 \times 20 = 1000 \text{ J}$. At the bottom, all $E_p$ converts to $E_k$: $E_k = 1000 \text{ J}$ $\frac{1}{2}mv^2 = 1000$ $\frac{1}{2} \times 5 \times v^2 = 1000$ $2.5v^2 = 1000$ $v^2 = \frac{1000}{2.5} = 400$ $v = \sqrt{400} = 20 \text{ m/s}$

Since energy is conserved, the potential energy at the highest point is equal to the kinetic energy at the lowest point.