Conservation of energy

The law of conservation of energy states that the total energy in an isolated system is constant. Energy is not created or destroyed; it changes form. For our purposes, total energy is the sum of kinetic and potential energies.

When a ball is thrown upwards, it is gaining potential energy and losing kinetic energy joule for joule. When it starts falling back down, the trade occurs in reverse. Mathematically,

$\mathrm{\Delta }{E}_{\text{k}}+\mathrm{\Delta }{E}_{\text{g}}=0\text{.}$

When we substitute those energy changes, we can rearrange to get

$v_2^2-v_1^2=-2g\mathrm{\Delta }h\text{.}$

This looks very similar to the eighth and final equation that we derived in the section on the equations of motion of the first unit.

If we were instead considering a situation where kinetic energy was converted to elastic energy (a ball hits a spring and compresses it), our equation would be $\mathrm{\Delta }{E}_{\text{k}}+\mathrm{\Delta }{E}_{\text{e}}=0\text{.}$

Example

A 2.5 kg block is dropped from 7.5 m above the floor. What is its speed as it hits the floor?

We can used the equation that I just mentioned:

$v_2^2-v_1^2=-2g\mathrm{\Delta }h\text{.}$

Solving for $v_2\text{,}$ we have

$v_2=\sqrt{v_1^2-2g\mathrm{\Delta }h}=\sqrt{0-2(9.80\text{N/kg})(-7.5\text{m})}\approx 12.124\text{,}$

therefore the final speed is 12 m/s.