{"id":28590,"date":"2024-09-16T13:00:53","date_gmt":"2024-09-16T13:00:53","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28590"},"modified":"2025-09-18T00:41:59","modified_gmt":"2025-09-18T00:41:59","slug":"work-and-mechanical-energy","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/work-and-mechanical-energy\/","title":{"rendered":"Work and Mechanical Energy"},"content":{"rendered":"<p><center><\/p>\n<h1>Work and Mechanical Energy<\/h1>\n<p><em><strong>Summary:<\/strong><br \/>\nWe will dedicate this series of classes to studying the relationship between work, energy, and some of its forms: kinetic and potential. From these analyses, we will calculate the stopping distance of a mobile that starts with an initial speed and travels until it stops. We will also simplify the study of free fall.<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Learning Objectives<\/strong><br \/>\nAt the end of this class, the student will be able to:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Understand<\/strong> the relationship between work and kinetic energy.<\/li>\n<li><strong>Apply<\/strong> the Living Forces Theorem to understand the variation of kinetic energy.<\/li>\n<li><strong>Explain<\/strong> the concept of potential energy and its relationship with the position of an object in a gravitational field.<\/li>\n<\/ol>\n<p><strong>INDEX<\/strong><br \/>\n<a href=\"#1\">Work and Kinetic Energy<\/a><br \/>\n<a href=\"#11\">The relationship between work and kinetic energy<\/a><br \/>\n<a href=\"#12\">Kinetic Energy and Braking Distance<\/a><br \/>\n<a href=\"#2\">Work and Potential Energy<\/a><br \/>\n<a href=\"#21\">Potential Energy and Free Fall<\/a><\/p>\n<p><\/center><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/_q-5C4FQOaU\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Work and Kinetic Energy<\/h2>\n<p style=\"text-align:justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_q-5C4FQOaU&amp;t=130s\" target=\"_blank\" rel=\"noopener\"><strong>When we apply a force on the block,<\/strong><\/a> we have the feeling that we add something to the system; intuitively, we say that \u00abwe are adding energy,\u00bb although this is not necessarily true. If the force is not enough to break the static friction, then it is not possible to say whether the block has changed in any way. Certainly, we are adding something to the system, an \u00abeffort,\u00bb so to speak, but it is also true that the static friction returns that \u00abeffort\u00bb in an opposite and complementary direction. However, when the static friction is broken, then it is possible to distinguish a change in the system: it is now in a different state of motion. To produce that new state of motion, we must add \u00absomething new\u00bb to the system: that \u00absomething\u00bb is what we call kinetic energy.<\/p>\n<p style=\"text-align:justify;\">Now we have two physical quantities that describe the state of a physical object or system: the linear momentum, which we know well and represents the state of motion, and the kinetic energy that we will begin to study and which, for now, <em>will represent what we have had to add to take the body or system from rest to that state of motion.<\/em><\/p>\n<p><a name=\"11\"><\/a><\/p>\n<h3>The relationship between work and kinetic energy<\/h3>\n<p style=\"text-align:justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_q-5C4FQOaU&amp;t=865s\" target=\"_blank\" rel=\"noopener\"><strong>Let&#8217;s go back to the block and the force<\/strong><\/a> that sets it in motion. If the force does not produce movement, then we say that it has not added anything to the system, while if it does, we say that it adds <strong>kinetic energy.<\/strong> When movement occurs, the body necessarily follows a certain path, and while it does, the force will add energy. This action of adding or removing kinetic energy is called \u00abperforming mechanical work,\u00bb thus <strong>the mechanical work element <span class=\"katex-eq\" data-katex-display=\"false\">dW<\/span> is defined through the equation<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{lr}\n\n    dW =\\vec{F} \\cdot d\\vec{r} &amp; (1)\n\n    \\end{array}<\/span>\n<\/p>\n<p style=\"text-align:justify;\">where <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}<\/span> is the applied force and <span class=\"katex-eq\" data-katex-display=\"false\">d\\vec{r}<\/span> is the displacement element on which the force has acted. Since this force is applied to a body of mass <span class=\"katex-eq\" data-katex-display=\"false\">m,<\/span> using the <a href=\"https:\/\/toposuranos.com\/las-leyes-de-newton\/\" target=\"_blank\" rel=\"noopener\">second law of Newton<\/a> we can write<\/p>\n<p style=\"text-align:center\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{lr}\n\n    \\displaystyle\\vec{F} =\\frac{d\\vec{p}}{dt} = m\\frac{d\\vec{v}}{dt} &amp; (2)\n\n    \\end{array}<\/span>\n<p style=\"text-align:justify;\">Thus, from equations (1) and (2) we will have<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{llr}\n\n    dW &amp; \\displaystyle =m\\frac{d\\vec{v}}{dt} \\cdot d\\vec{r} = m\\frac{d\\vec{r}}{dt} \\cdot d\\vec{v} = m\\vec{v} \\cdot d\\vec{v} &amp; (3)\n\n    \\end{array}<\/span>\n<p style=\"text-align:justify;\">Integrating this last expression to obtain the total work done, we will have:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{llr}\n\n    W &amp; = \\displaystyle {\\int_{i}^{f}} m\\vec{v} \\cdot d\\vec{v} = \\left.\\frac{1}{2}m \\|\\vec{v}\\|^2 \\right|_i^f &amp; \\\\ \\\\\n\n    &amp; \\displaystyle = \\frac{1}{2}m \\|\\vec{v}_f\\|^2 - \\frac{1}{2}m\\|\\vec{v}_i\\|^2 &amp; (4)\n\n    \\end{array}<\/span>\n<p style=\"text-align:justify;\">From this reasoning, we can see that mechanical work is equivalent to a difference of the same magnitude in two different states, one corresponding to a final state and another to the initial one. Such magnitude corresponds to what we have had to add (or remove) to change the state of motion, and it is what we call <strong>Kinetic Energy,<\/strong> and consequently, it is defined through the equation:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{llr}\n\n    E_{cin} &amp; \\displaystyle = \\frac{1}{2}m\\|\\vec{v}\\|^2 &amp; (5)\n\n    \\end{array}<\/span>\n<p style=\"text-align:justify;\">And therefore, we have that<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{llr}\n\n    W &amp; = \\Delta E_{cin} &amp;  (6)\n\n    \\end{array}<\/span>\n<p style=\"text-align:justify;\">This is what is known as the <strong>Living Forces Theorem.<\/strong><\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"background-color: #ddddff;\">\n<h4 style=\"color: #800000;\">Exercise<\/h4>\n<p style=\"text-align:justify;\"><strong>a) <a href=\"https:\/\/www.youtube.com\/watch?v=_q-5C4FQOaU&amp;t=1148s\" target=\"_blank\" rel=\"noopener\">A vehicle like the KIA Rio 5 has an approximate weight<\/a><\/strong> of 1,580 kg. Imagine you drive this vehicle in the city at a moderate speed of 50 km\/h and encounter a red light. To stop, the vehicle must dissipate that energy through the braking system. Then, to regain speed, it must obtain energy from the fuel through the engine. If the light had been green, there would have been no need to brake, saving energy. Calculate the amount of energy saved if the light is green.<\/p>\n<p>    <strong>b)<\/strong> Repeat the calculations from the previous section, but now assuming you drive at a less moderate speed of 70 km\/h. Express the extra energy expenditure as a percentage.<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/dGK-y2Pqu74\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"12\"><\/a><\/p>\n<h3>Kinetic Energy and Braking Distance<\/h3>\n<p style=\"text-align:justify;\">A very common mistake made by motorists is to intuit that the braking distance is directly proportional to speed: if we double the speed, the braking distance also doubles. In this section, we will review the error behind that intuition and demonstrate that, in reality, the braking distance is proportional to the square of the speed.<\/p>\n<p style=\"text-align:justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=dGK-y2Pqu74&amp;t=113s\" target=\"_blank\" rel=\"noopener\"><strong>Let&#8217;s suppose we have a block of mass<\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> moving on a horizontal plane with initial speed <span class=\"katex-eq\" data-katex-display=\"false\">v_i\\hat{x}<\/span> and where there is a kinetic friction coefficient <span class=\"katex-eq\" data-katex-display=\"false\">\\mu_c.<\/span> From this, we see that there is a friction force opposite to the motion <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{roce}=-\\mu_c mg\\hat{x}<\/span> that will act until the body stops after traveling the braking distance <span class=\"katex-eq\" data-katex-display=\"false\">x_{fre}.<\/span> The work done by this force is given by:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n    \\displaystyle W_{roce}= \\int_{0}^{x_{fre}} \\vec{F}_{roce} \\cdot d\\vec{l} = \\int_{0}^{x_{fre}} -\\mu_c mg dx = -\\mu_cmgx_{fre}\n\n    <\/span>\n<p style=\"text-align:justify;\">On the other hand, the variation of the kinetic energy of a body that starts with initial speed <span class=\"katex-eq\" data-katex-display=\"false\">v_i<\/span> and reaches rest with final velocity <span class=\"katex-eq\" data-katex-display=\"false\">v_f=0<\/span> is:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\Delta E_{cin}= \\frac{1}{2}m (\\underbrace{\\color{red}{v_f^2}}_{= 0} - v_i^2) = - \\frac{1}{2}mv_i^2\n\n    <\/span>\n<p style=\"text-align:justify;\">So that if all the kinetic energy is dissipated by friction until bringing the body to rest, we will have that:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\begin{array}{rrl}\n\n    &amp; W_{roce} &amp; \\displaystyle = \\Delta E_{cin} \\\\ \\\\\n\n \\equiv   &amp; -\\mu_cmg x_{fre} &amp; \\displaystyle = - \\frac{1}{2}mv_i^2 \\\\ \\\\\n\n \\equiv   &amp; x_{fre} &amp; = \\frac{1}{2} \\frac{v_i^2}{\\mu_c g}\n\n    \\end{array}<\/span>\n<p style=\"text-align:justify;\">From this, we have what we wanted to demonstrate, that the braking distance is proportional to the square of the speed.<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"background-color: #ddddff;\">\n<h4 style=\"color: #800000;\">Exercise<\/h4>\n<p style=\"text-align:justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=dGK-y2Pqu74&amp;t=510s\" target=\"_blank\" rel=\"noopener\"><strong>A block of 300[kg] moves at a speed<\/strong><\/a> of 15[km\/h] on a horizontal plane. If there is kinetic friction <span class=\"katex-eq\" data-katex-display=\"false\">\\mu_c=0.67<\/span> between the horizontal plane and the block, calculate the distance the block travels until it stops completely.<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/GXxkhk00quw\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Work and Potential Energy<\/h2>\n<p style=\"text-align:justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=GXxkhk00quw&amp;t=6s\" target=\"_blank\" rel=\"noopener\"><strong>Let&#8217;s imagine we have a body<\/strong><\/a> of mass <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> that falls from a height <span class=\"katex-eq\" data-katex-display=\"false\">h_i<\/span> to a final height <span class=\"katex-eq\" data-katex-display=\"false\">h_f<\/span> (with <span class=\"katex-eq\" data-katex-display=\"false\">h_f \\leq h_i<\/span>). Then we will have that the work done by the gravitational force is in the form:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle W_g = \\int_{h_i}^{h_f} \\vec{F}_g \\cdot d\\vec{l} = \\int_{h_i}^{h_f} -mgdz = -mg(h_f - h_i)<\/span>\n<p style=\"text-align:justify;\">From this, we have that if the initial height is <span class=\"katex-eq\" data-katex-display=\"false\">h_i = h<\/span> and the final height is at ground level <span class=\"katex-eq\" data-katex-display=\"false\">h_f = 0<\/span>, then we will have:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle W_g = -mg(0 - h) = mgh<\/span>\n<p style=\"text-align:justify;\">With this, we have that when a body falls from a height <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span>, energy associated with its relative position in space is released. This energy is what we call <strong>Potential Energy.<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\begin{array}{rr}{E_{pot} = mgh} &amp; (7)\\end{array}<\/span>\n<p><a name=\"21\"><\/a><\/p>\n<h3>Potential Energy and Free Fall<\/h3>\n<p style=\"text-align:justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=GXxkhk00quw&amp;t=329s\" target=\"_blank\" rel=\"noopener\"><strong>Let&#8217;s recall the problem of<\/strong><\/a> <a href=\"https:\/\/toposuranos.com\/movimiento-rectilineo-uniformemente-acelerado-mrua\/\" rel=\"noopener\" target=\"_blank\">free fall<\/a>. Using now the potential and kinetic energy we have studied, we can now find the falling speed with a much simpler procedure. Energy is another of those physical magnitudes that have the characteristic of being conservative; that is, it is not created or destroyed, only transformed. When we have a body set at a height <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span> from the ground, when it falls by the action of gravity until it reaches the ground, its potential energy does not disappear but transforms into another form of energy: kinetic energy, so we will have:<\/p>\n<p style=\"text-align:center\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n    E_{pot,inicial} &amp; = E_{cin, final} \\\\ \\\\ mgh &amp;\\displaystyle = \\frac{1}{2}mv^2 \\\\ \\\\ v^2 &amp; = 2gh \\\\ \\\\ v&amp; =\\sqrt{2gh}\\end{array}<\/span>\n<table>\n<tbody>\n<tr>\n<td style=\"background-color: #ddddff;\">\n<h4 style=\"color: #800000;\">Exercise<\/h4>\n<p>    <a href=\"https:\/\/www.youtube.com\/watch?v=GXxkhk00quw&amp;t=443s\" target=\"_blank\" rel=\"noopener\"><strong>A roller coaster has its starting point<\/strong><\/a> at a height of 150[m] from the ground. If the cart moves without friction on the roller coaster rails and starts from rest, calculate the speed when it is at a height of:<\/p>\n<p>    <strong>a)<\/strong> 90[m] from the ground.<\/p>\n<p>    <strong>b)<\/strong> 50[m] from the ground.<\/p>\n<p>    <strong>c)<\/strong> 10[m] from the ground.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>Work and Mechanical Energy Summary: We will dedicate this series of classes to studying the relationship between work, energy, and some of its forms: kinetic and potential. From these analyses, we will calculate the stopping distance of a mobile that starts with an initial speed and travels until it stops. We will also simplify the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28589,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":7,"footnotes":""},"categories":[651,635],"tags":[],"class_list":["post-28590","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fundamentals-of-mechanics","category-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Work and Mechanical Energy - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Work and energy are related: work is when a force moves something and gives it energy. Energy is what allows things to move or change, like when something is in motion (kinetic energy) or in a high position (potential energy). 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Energy is what allows things to move or change, like when something is in motion (kinetic energy) or in a high position (potential energy). 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