Work-energy principle

The work done on a body by external forces is equal to the change in total mechanical energy of that body. Hence the diffrence in mechanical energy would be the work done against resistive forces.

Example

The following system in projected up the slope with speed of $6.25\text{ ms}^{-1}$. The surface has a coefficient of friction of 0.22. Find the objects speed as it passes back though its initial point.

Formulate the initial system

$$\begin{align} \text{Initial M.E} &= \text{G.P.E} + \text{K.E}\\ &= (5.5 \times g \times 0) + \left(\frac{1}{2} \times 5.5 \times 6.25^2 \right)\\ &= 107.421\text{ J} \end{align}$$

Formulate the system at instantaneous rest

$$\begin{align} \text{Rest M.E.} &= \text{G.P.E} + \text{K.E}\\ &= (5.5 \times g \times x\sin(17^{\circ})) + \left(\frac{1}{2} \times 5.5 \times 0^2 \right)\\ &= 15.759x \end{align}$$

Formulate the final system

$$\begin{align} \text{Final M.E.} &= \text{G.P.E} + \text{K.E}\\ &= (5.5 \times 9.8 \times 0) + \left(\frac{1}{2} \times 5.5 \times v^2\right)\\ &= 2.75v^2 \text{ J} \end{align}$$

Calculate the work done by friction (on the way up)

$$\begin{align} W &= Fd\\ &= \mu R \times x\\ &= 0.22 \times 5.5gcos(17^{\circ}) \times x\\ &= 11.340x \text{ J} \end{align}$$

Find $x$

$$\begin{align} \text{Initial M.E.} - \text{Rest M.E.} &= \text{Work against friction}\\ 107.421 - 15.759x &= 11.340x\\ 107.421 &= 27.099x\\ x &= 3.964\\ &= 3.96 \text{ m} \end{align}$$

Find v

$$\begin{align} \text{Initial M.E.} - \text{Final M.E.} &= 2 \times \text{Work against friction}\\ 107.421 - 2.75v^2 &= 2 \times 11.34 \times 3.96 \\ &= 89.904\\ -2.75v^2 &= -17.517\\ v^2 &= 6.370\\ v &= 2.52 \text{ms}^{-1} \end{align}$$