<h2>Net Work</h2>
<p>We have only looked at a single force acting on an object. Sometimes more than one force acts at the same time. We call the work done after taking all the forces into account the <strong>net work done</strong>. If there is only one force acting then the work it does, if any, is the net work done. In this case there are two equivalent approaches we can adopt to finding the net work done on the object. We can:</p>
<ul>
<li><strong>Approach 1:</strong> calculate the work done by <strong>each force individually</strong> and then sum them <strong>taking the signs into account</strong>. If one force does positive work and another does the same amount of work but it is negative then they cancel out.</li>
<li><strong>Approach 2:</strong> calculate the resultant force from all the forces acting and calculate the work done using the resultant force. This will be equivalent to Approach 1. If the resultant force parallel to the direction of motion is zero, no net work will be done.</li>
</ul>
<p>Remember that work done tells you about the energy transfer to or from an object by means of a force. That is why we can have zero net work done even if multiple large forces are acting on an object. Forces that result in positive work increase the energy of the object, forces that result in negative work reduce the energy of an object. If as much energy is transferred to an object as is transferred away then the final result is that the object gains no energy overall.</p>
<div class="ns-school-emp">
<h2>Example: Approach 1, Calculating the Net Work on a Car</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>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.</p>
<img alt="" src="https://data.ulearngo.com/assets/3346f6ec-533e-4ab6-aeca-58d6ec042362"/>
<p>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.</p>
<h3>Solution</h3>
</div>
<div>
<h3>Step 1: Analyse the question to determine what information is provided</h3>
<ul>
<li>
<p>The force applied is \(F_{\text{applied}}\)=\(\text{300}\) \(\text{N}\) forwards.</p>
</li>
<li>
<p>The force of friction is \(F_{\text{friction}}\)=\(\text{100}\) \(\text{N}\) opposite to the direction of motion.</p>
</li>
<li>
<p>The distance moved is \(\Delta x\) = \(\text{20}\) \(\text{m}\).</p>
</li>
<li>
<p>The applied force and distance moved are in the same plane so we can calculate the work done by the applied forward force and the work done by the force of friction backwards.</p>
</li>
</ul>
<p>These quantities are all in the correct units, so no unit conversions are required.</p>
</div>
<div>
<h3>Step 2: Analyse the question to determine what is being asked</h3>
<ul>
<li>
<p>We are asked to find the net work done on the car. We know from the definition that work done is \(W={F}\)\(\Delta x\) \(\cos\theta\)</p>
</li>
</ul>
</div>
<div>
<h3>Step 3: Next we calculate the work done by each force.</h3>

\begin{align*} W_{\text{applied}}&amp; = {F}_{\text{applied}}\Delta x\cos\theta\\ &amp; = \left(300\right)\left(20\right) \left(\cos 0\right)\\ &amp; = \left(300\right)\left(20\right) \left(1\right)\\ &amp; = \text{6 000}\text{ J} \end{align*}\begin{align*} W_{\text{friction}}&amp; = {F}_{\text{friction}}\Delta x\cos\theta\\ &amp; = \left(100\right)\left(20\right) \left(\cos 180\right)\\ &amp; = \left(100\right)\left(20\right) \left(-1\right)\\ &amp; = -\text{2 000}\text{ J} \end{align*}\begin{align*} {W}_{\text{net}}&amp; = {W}_{\text{applied}} + {W}_{\text{friction}} \\ &amp; = \left(6000\right) + \left(-2000\right) \\ &amp; = \text{4 000}\text{ J} \end{align*}

<p>The answer shown in this worked example shows that although energy has been lost by the car to friction, the total work done on the car has resulted in a net energy gain. This can be seen by the positive answer.</p>
</div>
</div>
</div>

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Net Work

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Explore the concepts of mechanical energy, potential and kinetic energy, work, power, and conservation of energy through examples like pendulum, roller coaster, and inclined plane.


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