Begin with a kinematic equation for constant acceleration and Newton's second law. $v^2 = v{_0}^2 +2ax $ $a = \frac{F}{m}$ = constant $v^2 = {v_0}^2 + 2 \frac{F}{m} x$ Substitute Newton's second into the kinematic equation. $\frac{1}{2}mv^2 - \frac{1}{2}m{v_0}^2 = {F} \cdot x$ Multiply both sides by $\frac{m}{2}$ $KE = \frac{1}{2}mv^2$ $W = F \cdot x$ $\therefore \Delta KE = W$ --- Here is another way: $W = F \cdot x$ Substitute F = ma into the Work equation $W = ma \cdot x$ $v^2 = v_0^2 +2ax$ Rearrange and solve for a $a = \frac{v^2-{v_0}^2}{2x}$ $W = m (\frac{v^2-{v_0}^2}{2x}) \cdot x$ Substitute acceleration from kinematics into Work equation $W = m (\frac{v^2}{2} -\frac{{v_0}^2}{2} )$ $W = \frac{1}{2}mv^2 - \frac{1}{2}m{v_0}^2$ $W = \Delta KE$ Cancel the x **Work Energy Theorem** (General Statement) This shows work is a transfer of energy. Work is NOT energy. --- ### Related Topics --- [[Home|Home]] | [[Energy]] | [[Mechanical Energy]] | [[Kinetic Energy]] | [[Work-Energy Theorem]] | [[Potential Energy]] | [[Law of Conservation of Energy]] | [[Power]]