<h2>Examples on Newton's Second Law of Motion</h2>
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<h3>Example: What Acceleration Can a Person Produce when Pushing a Lawn Mower?</h3>
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<p>Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the mower is 24 kg. What is its acceleration?</p>
<img alt="A man pushing a lawnmower to the right. A red vector above the lawnmower is pointing to the right and labeled F sub net." src="https://data.ulearngo.com/assets/ddbef280-0225-4603-91e1-10fb3850b57f"/>The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right?

<p><strong>Strategy</strong></p>
<p>Since \({\mathbf{\text{F}}}_{\text{net}}\) and \(m\) are given, the acceleration can be calculated directly from Newton's second law as stated in \({\mathbf{\text{F}}}_{\text{net}}=m\mathbf{\text{a}}\).</p>
<p><strong>Solution</strong></p>
<p>The magnitude of the acceleration \(a\) is \(a=\frac{{F}_{\text{net}}}{m}\). Entering known values gives</p>
<p>\(a=\frac{\text{51 N}}{\text{24 kg}}\)</p>
<p>Substituting the units \(\text{kg}\cdot {\text{m/s}}^{2}\) for N yields</p>
<p>\(a=\frac{\text{51 kg}\cdot {\text{m/s}}^{2}}{\text{24 kg}}=\text{2.1 m}{\text{/s}}^{2}.\)</p>
<p><strong>Discussion</strong></p>
<p>The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person's top speed would soon be reached.</p>
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<h3>Example: What Rocket Thrust Accelerates This Sled?</h3>
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<p>Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust \(\mathbf{\text{T}}\), for the four-rocket propulsion system shown in the figure below. The sled's initial acceleration is \(\text{49}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2},\) the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.</p>
<div class="wp-caption alignnone" style="width: 400px"><img alt="A sled is shown with four rockets, each producing the same thrust, represented by equal length arrows labeled as vector T pushing the sled toward the right. Friction force is represented by an arrow labeled as vector f pointing toward the left on the sled. The weight of the sled is represented by an arrow labeled as vector W, shown pointing downward, and the normal force is represented by an arrow labeled as vector N having the same length as W acting upward on the sled. A free-body diagram is also shown for the situation. Four arrows of equal length representing vector T point toward the right, a vector f represented by a smaller arrow points left, vector N is an arrow pointing upward, and the weight W is an arrow of equal length pointing downward." height="326" src="https://data.ulearngo.com/assets/aef76506-e99c-4eea-aa0e-1c1b8e409a76" width="400"/><p class="wp-caption-text">A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust \(\mathbf{\text{T}}\). As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force \(\mathbf{\text{N}}\) on the system that is equal in magnitude and opposite in direction to its weight, \(\mathbf{\text{w}}\). The system here is the sled, its rockets, and rider, so none of the forces <em>between</em> these objects are considered. The arrow representing friction (\(\mathbf{\text{f}}\)) is drawn larger than scale.</p></div>
<p><strong>Strategy</strong></p>
<p>Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.</p>
<p><strong>Solution</strong></p>
<p>Since acceleration, mass, and the force of friction are given, we start with Newton's second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with</p>
<p>\({F}_{\text{net}}=\text{ma},\)</p>
<p>where \({F}_{\text{net}}\) is the net force along the horizontal direction. We can see from the figure above that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is</p>

\({F}_{\text{net}}=4T-f.\)

<p>Substituting this into Newton's second law gives</p>

\({F}_{\text{net}}=\text{ma}=4T-f.\)

<p>Using a little algebra, we solve for the total thrust 4<em>T</em>:</p>

\(4T=\text{ma}+f.\)

<p>Substituting known values yields</p>

\(4T=\text{ma}+f=\left(\text{2100 kg}\right)\left({\text{49 m/s}}^{2}\right)+\text{650 N}.\)

<p>So the total thrust is</p>
<p>\(4T=1.0×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N},\)</p>
<p>and the individual thrusts are</p>
<p>\(T=\frac{1.0×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}}{4}=2.6×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}.\)</p>
<p><strong>Discussion</strong></p>
<p>The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections.</p>
<p>Speeds of 1000 km/h were obtained, with accelerations of 45 \(g\)'s. (Recall that \(g\), the acceleration due to gravity, is \(9\text{.}{\text{80 m/s}}^{2}\). When we say that an acceleration is 45 \(g\)'s, it is \(\text{45}×9\text{.}{\text{80 m/s}}^{2}\), which is approximately \({\text{440 m/s}}^{2}\).)</p>
<p>While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.</p>
<p>Newton's second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. The next section introduces the third and final law of motion.</p>
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Examples on Newton's Second Law of Motion

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Discover the development of the force concept, explore Newton&#8217;s first, second, and third laws of motion, and learn about problem-solving strategies and further applications in this tutorial on dynamics and the four basic forces.


Examples on Newton's Second Law of Motion

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