<h2 class="title">Work</h2>
<p>We cover different topics in different tutorials but that doesn't mean that they are not related. In fact, it is very important to note that all of the different topics related to mechanics (forces, mechanical energy, momentum, rectilinear motion) actually form a <strong>consistent</strong> picture of the same physical system. There have been examples where we've shown the same results using two methods, for example determining speed or velocity using equations of motion or conservation of mechanical energy. Learning about work will help us tie everything we've learnt about previously together.</p>
<p>Work will allow us to connect energy transfer to forces, which we have already linked to momentum and the equations of motion. When a force tends to act in or against the direction of motion of an object we say that the force is doing work on the object. Specifically, work is defined mathematically in terms of the force and the displacement of the object.</p>
<div class="ns-school-emp">
<h3>Definition: Work</h3>
<div class="ns-school-defn">
<p>When a force acts in or against the direction of motion of an object, work is done on the object.</p>

\[W=F\Delta x\cos\theta\]</div>
</div>
<p>This means that in order for work to be done, an object must have its position changed by an amount \(\Delta \vec{x}\) while a force, \(\vec{F}\), is acting on it such that there is some non-zero component of the force in the direction of the displacement. Work is calculated as:</p>
<p>\[W=F\Delta x\cos\theta\]</p>
<p>Where:</p>
<ul>
<li>\(W\) = work in joules (J)</li>
<li>\(F\) = magnitude of force for which we are calculating work in newtons (N)</li>
<li>\(\Delta x\) = magnitude of displacement in metres (m)</li>
<li>\(\theta\) = angle between the force and the direction of displacement</li>
</ul>
<p>Note that:</p>
<ul>
<li>\(F\cos\theta\) is the magnitude of the component of \(\vec{F}\) in the direcion of \(\Delta \vec{x}\). If \(\theta &gt; 90^{o}\) then the component is parallel to the direction of displacement but point in the opposite direction and the force is opposing the motion.</li>
<li>We will deal with constant forces not forces that are continuously changing in this tutorial. All of the principles we have covered are applicable when forces are not constant.</li>
</ul>
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<td><img alt="7bd581d481c6c6498b2cffc5e69a4759.png" src="https://data.ulearngo.com/assets/20c6e828-5821-4e4e-bbb5-ed9176c3be72"/></td>
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<p>The force \(\vec{F}\) causes the object to be displaced by \(\Delta \vec{x}\) at angle θ.</p>
<p><strong>Important:</strong> \(\cos\theta\) tells you the relative direciton of the force and the displacment which is important. If the component of the force along the direction of the displacement is opposite in direction to the displacement then the sign of the displacement vector and force vector will be different. This is regardless of which direction was chosen as a positive direction.</p>
<p>Let us look at some examples to understand this properly. In the images below the grey dot represents an object. A force, \(\vec{F}\), acts on the object. The object moves through a displacement, \(\Delta \vec{x}\). What is the sign of the work done in each case?</p>
<p>At \(\theta=90^\circ\) the magnitude of the component of the force in the direction of the displacement is zero so <strong>no work is done</strong> as \(\cos(90)=0\).</p>
<p><img alt="e8c619736c65bc054dee0a88ce0bc724.png" src="https://data.ulearngo.com/assets/fdbf96d3-a98f-441d-b69d-cc95245cdc9f"/></p>
<p>At \(\theta&gt;90^\circ\) the magnitude of the component of the force along the line of the displacement is opposite in direction to the displacement so <strong>negative work is done</strong>, \(\cos\theta &lt; 0\).</p>
<p><img alt="7e4d7f08ed235810217b22936dfa9126.png" src="https://data.ulearngo.com/assets/88ba9050-b6dd-4763-9b1f-0437f91955bc"/></p>
<p>It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work.</p>
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<p>The force \(\vec{F}\) causes the object to be displaced by \(\Delta \vec{x}\) at angle θ.</p>
<p>It is only the direction of the force on the object that matters and not the direction from the source of the force to the object. In the figure below, both powerlifters are exerting an upwards force on the weights. On the left the weight is being pulled upwards and on the right it is being pushed upwards.</p>
<div class="wp-caption alignnone" style="width: 800px"><img alt="" height="531" src="https://data.ulearngo.com/assets/d0002409-5b91-499b-bcf6-6c74529852e2" width="800"/><p class="wp-caption-text">A powerlifter deadlifts a weight.</p></div> <div class="wp-caption alignnone" style="width: 800px"><img alt="Bench press" height="538" src="https://data.ulearngo.com/assets/19231928-0a0c-4469-b43a-d04268220c95" width="800"/><p class="wp-caption-text">A powerlifter benchpresses a weight.</p></div>
<p>Weight lifting is a good context in which to think about work because it helps to identify misconceptions introduced by everyday use of the term `work'. In the two cases in the figure above everyone would describe moving the weights upwards as very hard work. From a physics perspective, as the powerlifters lift the weight they are exerting a force in the direction of the displacement so positive work is done on the weights.</p>
<p>Consider the strongman walking in the figure below. He carries two very heavy sleds as far as he can in a competition. What work is the man doing on the sleds and why?</p>
<div class="wp-caption alignnone" style="width: 800px"><img alt="" height="1200" src="https://data.ulearngo.com/assets/443e0a37-21cf-46f7-9330-59676ec7277d" width="800"/><p class="wp-caption-text">A strongman carries heavy sleds as far as possible in a competition.</p></div>
<p>Most people would say he is working very hard because the sleds are heavy to carry but from a physics perspective he is doing no work on the sleds. The reason that he does no work is because the force he exerts is directly upwards to balance the force of gravity and the displacement is in the horizontal direction. Therefore there is no component of the force in the direction of displacement (\(\theta=90^\circ\)) and no work done.</p>
<p>His muscles do need to use their energy reserves to maintain the force to balance gravity. That does not result in energy transfer to the sleds.</p>

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