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,

ΔEk+ΔEg=0\displaystyle \Delta{} E_{\text{k}} + \Delta{} E_{\text{g}} = 0.

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

v22v12=2gΔh\displaystyle v_{2}^{2} - v_{1}^{2} = - 2 g{\Delta{}} h.

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 ΔEk+ΔEe=0\displaystyle \Delta{} E_{\text{k}} + \Delta{} E_{\text{e}} = 0.


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:

v22v12=2gΔh\displaystyle v_{2}^{2} - v_{1}^{2} = - 2 g{\Delta{}} h.

Solving for v2\displaystyle v_{2}, we have

v2=v122gΔh=02(9.80N/kg)(7.5m)12.124\displaystyle v_{2} = \sqrt{v_{1}^{2} - 2 g{\Delta{}} h} = \sqrt{0 - 2 \left ( 9.80 \, \text{N/kg} \right ) \left ( - 7.5 \, \text{m} \right )} \approx 12.124,

therefore the final speed is 12 m/s.