Net Work Continued

The same car is now accelerating forward, but friction is working against the motion of the car. A force of \(\text{300}\) \(\text{N}\) is applied forward on the car while it is travelling \(\text{20}\) \(\text{m}\) forward. A frictional force of \(\text{100}\) \(\text{N}\) acts to oppose the motion. Calculate the net work done on the car.

Only forces with a component in the plane of motion are shown on the diagram. No work is done by \({F}_{g}\) or \({F}_{\text{normal}}\) as they act perpendicular to the direction of motion. The net force acting in the plane of motion will be calculated using the non-perpendicular forces.

Step 1: Analyse the question to determine what information is provided

• The force applied is \(\vec{F}_{\text{applied}}\)=\(\text{300}\) \(\text{N}\) forwards.

• The force of friction is \(\vec{F}_{\text{friction}}\)=\(\text{100}\) \(\text{N}\) backwards.

• The distance moved is \(\Delta x\) = \(\text{20}\) \(\text{m}\).

• The applied forces \(\vec{F}_{\text{applied}}= \text{300}\text{ N}\) and the force of friction \(\vec{F}_{\text{friction}}= \text{100}\text{ N}\) are in the same plane as the distance moved. Therefore, we can add the vectors. As vectors require direction, we will say that forward is positive and therefore backward is negative. Note, the force of friction is acting at \({180}^{0}\) i.e. backwards and so is acting in the opposite vector direction i.e. negative.

These quantities are all in the correct units, so no unit conversions are required.

Step 2: Analyse the question to determine what is being asked

• We are asked to find the net work done on the car. We know from the definition that work done is \(W_{\text{net}}={F}_{\text{net}}\)\(\Delta x \cos \theta\)

Step 3: We calculate the net force acting on the car, and we convert this into net work.

First we draw the force diagram:

Let forwards (to the left in the picture) be positive. We know that the motion of the car is in the horizontal direction so we can neglect the force due to gravity, \(\vec{F}_g\), and the normal force, \(\vec{N}\). Note: if the car were on a slope we would need to calculate the component of gravity parallel to the slope.

\begin{align*} \vec{F}_{\text{net}}& = \vec{F}_{\text{applied}} + \vec{F}_{\text{friction}}\\ & = \left(+300\right) + \left(-100\right)\\ \vec{F}_{\text{net}}& = \text{200}\text{ N}~\text{forwards} \end{align*}

\(\vec{F}_{\text{net}}\) is pointing in the same direction as the displacement, therefore the angle between the force and displacement is \(\theta=0^\circ\).

\begin{align*} {W}_{\text{net}}& = {F}_{\text{net}}\Delta x \cos \theta \\ & = \left(200\right)\left(20\right)\cos\left(0\right)\\ & = \text{4 000}\text{ J} \end{align*}