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<h2>Example: Block on an Inclined Plane</h2>
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<h3>Question</h3>
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<p>A block of $\text{2}$ $\text{kg}$ is pulled up along a smooth incline of length $\text{10}$ $\text{m}$ and height $\text{5}$ $\text{m}$ by applying an non-conservative force. At the end of incline, the block is released from rest to slide down to the bottom. Find the</p>
<ol>
<li>work done by the non-conservative force,</li>
<li>the kinetic energy of the block at the end of round trip, and</li>
<li>the speed at the end of the round trip.</li>
</ol>
</div>
<div>
<h3>Step 1: Analyse what is given and what is required</h3>
<p>There are three forces on the block while going up:</p>
<ol>
<li>weight of the block, $\vec{F}_g=m\vec{g}$,</li>
<li>normal force, $\vec{N}$, applied by the block and</li>
<li>non-conservative force, $\vec{F}$.</li>
</ol>

On the other hand, there are only two forces while going down. The non-conservative force is absent in downward journey. The force diagram of the forces is shown here for upward motion of the block.<img alt="35c70d1b9b9aaaa98a23f5c40bcfc966.png" src="https://data.ulearngo.com/assets/38f3a3c2-737d-430c-a5d5-eaacdaf8f951"/></div>
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<h3>Step 2: Work done by non-conservative force during round trip</h3>
<p>The work done by the non-conservative force we will call $W_F$.</p>
<p>We have not been given a magnitude or direction for $\vec{F}$, all we do know is that it must result in the block moving up the slope.</p>
<p>We have represented the non-conservative force on the force diagram with an arbitrary vector. $\vec{F}$ acts only during upward journey. Note that the block is simply released at the end of the upward journey. We need to find the work done by the non-conservative force only during the upward journey.</p>

\[W_F=W_{F\text{ (up)}}+W_{F\text{ (down)}}=W_{F\text{ (up)}}+0=W_{F\text{ (up)}}\]

<p>The kinetic energies in the beginning and at the end of the motion up the slope are zero.</p>
<p>We can conclude that sum of the work done by all three forces is equal to zero during the upward motion. The change in kinetic energy is zero which implies that the net work done is zero.</p>

\begin{align*} W_{\text{net}} &amp; = W_{F(up)} + W_{g(up)} + W_{N(up)} \\ 0 &amp; = W_{F(up)} + W_{g(up)} + W_{N(up)} \end{align*}

<p>If we know the work done by the other two forces (normal force and gravity), then we can calculate the work done by the non-conservative force, $F$, as required.</p>
</div>
<div>
<h3>Step 3: Work done by normal force during upward motion</h3>
<p>The block moves up the slope, the normal force is perpendicular to the slope and, therefore, perpendicular to the direction of motion. Forces that are perpendicular to the direction of motion do no work.</p>

\begin{align*} 0 &amp; = W_{F\text{ (up)}} + W_{g\text{ (up)}} + W_{N\text{ (up)}} \\ 0 &amp; = W_{F\text{ (up)}} + W_{g\text{ (up)}} + (0)\\ W_{F\text{ (up)}} &amp; = - W_{g\text{ (up)}} \end{align*}

<p>Thus, we need to compute work done by the gravity in order to compute work by the non-conservative force.</p>
</div>
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<h3>Step 4: Work done by gravity during upward motion</h3>
<p>The component of the force due to gravity that is perpendicular to the slope, $\vec{F}_{gy}$, will do no work so we only need to consider the component parallel to the slope, $\vec{F}_{gx}$.</p>
<p>For the upward motion $\vec{F}_{gx}$ is in the opposite direction to the motion, therefore the angle between them is $\theta = 180°$, which means $\cos \theta = -1$, so we expect to see negative work is done.</p>
<p>The magnitude of $\vec{F}_{gx}$ is $F_{gx} = mg\sin\alpha$. So the work done by gravity during the upward motion is :</p>

\begin{align*} W_{g\text{ (up)}}&amp; = F_{gx} \Delta x \cos \theta \\ &amp; = mg \sin\alpha\cdot\Delta x \cos \theta\\ &amp;=(2)(\text{9.8})(\frac{5}{10})(10)\cos(180)\\ &amp;=−\text{98}\text{ J} \end{align*}

<div class="ns-school-defn2">
<h3>Important:</h3>
<p>Be careful not to be confused by which angle has been labelled $\alpha$ and which $\theta$. $\alpha$ is not the angle between the force and the direction of motion but the incline of the plane in this particular problem. It is important to understand which symbol represents which physical quantity in the equations you have learnt.</p>
</div>
<p>Hence, the work done by the non-conservative force during the round trip is:</p>

\begin{align*} W_F = W_{F\text{ (up)}} &amp; = - W_{g\text{ (up)}} \\ =−(−\text{98})\\ =\text{98}\text{ J} \end{align*}</div>
<div>
<h3>Step 5: Kinetic energy at the end of round trip</h3>
<p>The kinetic energy at the end of the upward motion was zero but it is not zero at the end of the entire downward motion.</p>
<p>We can use the work-energy theorem to analyse the whole motion:</p>

\begin{align*} W_{(round\ trip)} &amp; = E_{k,f} - E_{k,i} \\ &amp; = E_{k,f} - 0 \\ &amp; = E_{k,f} \end{align*}

<p>To determine the net work done, $W_{\text{ (round trip)}}$, we need to sum the work done by each force acting during the period. We have calculated the work done by $\vec{F}$ already and we know that there is no work done by the normal force.</p>
<p>The total work done during round trip by gravity is the sum of the work done during the upward motion (where the force is in the opposite direction to the motion) and the downward motion (where the force is in the same direction as the motion).</p>
<p>The distance over which the force acts is the same during the upward and downward motion and the magnitude of the force is the same. The only difference between the calculation for the work done during the upward and downward motion is the sign because of the change of direction of the motion. Therefore:</p>

\begin{align*} W_{g\text{ (round trip)}} &amp;=W_{g\text{ (up)}} + W_{g\text{ (down)}} \\ &amp; = ( mg\sin\alpha \cdot \Delta x \cos 180°) + (mg\sin\alpha \cdot \Delta x \cos 0°)\\ &amp; = ( −mg\sin\alpha \cdot \Delta x) + (mg\sin\alpha \cdot\Delta x)\\ &amp;= \text{0}\text{ J} \end{align*}

<p>Hence, the total work done during round trip is:</p>

\begin{align*} W_{(round\ trip)} &amp; = W_{F} + W_{g} + W_{N} \\ &amp; = W_{F} + W_{g(up)} + W_{N(up)} \\ &amp; = (\text{98})+ (0) + (0) \\ &amp; = \text{98}\text{ J} \end{align*}

<p>We can now use this in the equation for the work-energy theorem:</p>

\begin{align*} W_{(round\ trip)} &amp; = E_{k,f} \\ E_{k,f} &amp; =\text{98}\text{ J} \end{align*}</div>
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<h3>Step 6: Speed of the block</h3>
<p>We know the kinetic energy of the block therefore we can determine its speed:</p>

\begin{align*} E_{k,f} &amp;= \frac{1}{2}mv^2 \\ \frac{1}{2}mv^2&amp;= E_{k,f} \\ v &amp;=\pm\sqrt{\frac{2}{m}E_{k,f}} \\ &amp;=\pm\sqrt{\frac{2}{(2)}(\text{98})} \\ &amp;=\pm\text{9.89949493661} \\ &amp;=\text{9.90}\text{ m·s$^{-1}$} \end{align*}

<p>Note that the total work done during the upward motion is zero as the block is stationary at the beginning and at the end of the motion up the incline. The positive work done by the non-conservative force is cancelled by the fact that exactly the same amount of negative work is done by gravity. The net work comes from the work done during the downward motion by gravity. Net work done results in a change in kinetic energy as per the work-energy theorem.</p>
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Block on an Inclined Plane

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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.


Block on an Inclined Plane

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