{"id":230209,"date":"2025-06-09T05:20:16","date_gmt":"2025-06-09T05:20:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230209"},"modified":"2025-06-09T05:21:08","modified_gmt":"2025-06-09T05:21:08","slug":"the-mass-of-the-hanging-mass-is-m-and-the-mass-of-the-cart-is-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/09\/the-mass-of-the-hanging-mass-is-m-and-the-mass-of-the-cart-is-m\/","title":{"rendered":"The mass of the hanging mass is m and the mass of the cart is M."},"content":{"rendered":"\n<p>The mass of the hanging mass is m and the mass of the cart is M. You can assume that the floor over which the mass hangs has been chosen as the zero point of gravitational potential energy. If the hanging mass is at a height<br>above the floor, and has not yet been released, what type(s) of energy is\/are present in the system? NOTE: You can ignore any energies that are not relevant (will not change) as the mass is held, released, and allowed to fall. What is the final energy of the system after the weight has traveled a distance<br>? Leave your answer in symbolic terms. Answer in terms of<br>,<br>,<br>, and\/or<br>, and use<br>for the acceleration due to gravity.<\/p>\n\n\n\n<p>Assume that energy is conserved. What is the final velocity of the cart, based on your expressions for initial and final energy? Leave your answer in symbolic terms. Answer in terms of<br>,<br>, and\/or<br>, and use<br>for the acceleration due to gravity.<br>What is the acceleration of the cart-mass system to this point? Leave your answer in symbolic terms. Answer in terms of<br>,<br>, and\/or<br>, and use<br>for the acceleration due to gravity. HINT: How are velocity, acceleration, and traveled distance related to each other? Look at your Lab02 results\/derivations.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let\u2019s analyze the system step-by-step and derive symbolic expressions for:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial energy<\/li>\n\n\n\n<li>Final energy<\/li>\n\n\n\n<li>Final velocity<\/li>\n\n\n\n<li>Acceleration<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>System Overview<\/strong><\/h3>\n\n\n\n<p>We have:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A <strong>cart of mass<\/strong> MMM<\/li>\n\n\n\n<li>A <strong>hanging mass<\/strong> mmm<\/li>\n\n\n\n<li>The <strong>hanging mass<\/strong> is at height hhh above the floor (initially), and it&#8217;s connected to the cart via a pulley system<\/li>\n\n\n\n<li><strong>Distance fallen<\/strong> by the mass: ddd<\/li>\n\n\n\n<li><strong>Gravity<\/strong>: ggg<\/li>\n\n\n\n<li><strong>Assumption<\/strong>: No friction, no energy loss \u2014 <strong>conservation of mechanical energy<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Initial Energy<\/strong><\/h3>\n\n\n\n<p>At the start:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The cart is stationary: <strong>no kinetic energy<\/strong><\/li>\n\n\n\n<li>The hanging mass is stationary but elevated: <strong>gravitational potential energy<\/strong> is present<\/li>\n<\/ul>\n\n\n\n<p>Thus, total <strong>initial energy<\/strong> EiE_iEi\u200b:Ei=mghE_i = mghEi\u200b=mgh<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Final Energy<\/strong><\/h3>\n\n\n\n<p>After the hanging mass falls a vertical distance ddd:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The cart (mass MMM) and the hanging mass (mass mmm) both move with the same speed vvv<\/li>\n\n\n\n<li>The cart is moving horizontally, the hanging mass is falling vertically<\/li>\n\n\n\n<li>Final <strong>kinetic energy<\/strong> is:<\/li>\n<\/ul>\n\n\n\n<p>Kf=12Mv2+12mv2K_f = \\frac{1}{2}Mv^2 + \\frac{1}{2}mv^2Kf\u200b=21\u200bMv2+21\u200bmv2<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Final <strong>potential energy<\/strong> of hanging mass is:<\/li>\n<\/ul>\n\n\n\n<p>Uf=mg(h\u2212d)U_f = mg(h &#8211; d)Uf\u200b=mg(h\u2212d)<\/p>\n\n\n\n<p>So total <strong>final energy<\/strong> EfE_fEf\u200b:Ef=12Mv2+12mv2+mg(h\u2212d)E_f = \\frac{1}{2}Mv^2 + \\frac{1}{2}mv^2 + mg(h &#8211; d)Ef\u200b=21\u200bMv2+21\u200bmv2+mg(h\u2212d)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>3. Use Energy Conservation<\/strong><\/h3>\n\n\n\n<p>Ei=Ef\u21d2mgh=12Mv2+12mv2+mg(h\u2212d)E_i = E_f \\Rightarrow mgh = \\frac{1}{2}Mv^2 + \\frac{1}{2}mv^2 + mg(h &#8211; d)Ei\u200b=Ef\u200b\u21d2mgh=21\u200bMv2+21\u200bmv2+mg(h\u2212d)<\/p>\n\n\n\n<p>Cancel mghmghmgh from both sides:0=12Mv2+12mv2\u2212mgd0 = \\frac{1}{2}Mv^2 + \\frac{1}{2}mv^2 &#8211; mgd0=21\u200bMv2+21\u200bmv2\u2212mgd<\/p>\n\n\n\n<p>Multiply through by 2:0=Mv2+mv2\u22122mgd\u21d2v2(M+m)=2mgd\u21d2v=2mgdM+m0 = Mv^2 + mv^2 &#8211; 2mgd \\Rightarrow v^2(M + m) = 2mgd \\Rightarrow v = \\sqrt{\\frac{2mgd}{M + m}}0=Mv2+mv2\u22122mgd\u21d2v2(M+m)=2mgd\u21d2v=M+m2mgd\u200b\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>4. Acceleration of the System<\/strong><\/h3>\n\n\n\n<p>Use the kinematic relation:v2=2ad\u21d2a=v22dv^2 = 2ad \\Rightarrow a = \\frac{v^2}{2d}v2=2ad\u21d2a=2dv2\u200b<\/p>\n\n\n\n<p>Substitute v2=2mgdM+mv^2 = \\frac{2mgd}{M + m}v2=M+m2mgd\u200b:a=2mgd(M+m)\u22c52d=mgM+ma = \\frac{2mgd}{(M + m) \\cdot 2d} = \\frac{mg}{M + m}a=(M+m)\u22c52d2mgd\u200b=M+mmg\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answers<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Final velocity<\/strong>: v=2mgdM+mv = \\sqrt{\\frac{2mgd}{M + m}}v=M+m2mgd\u200b\u200b<\/li>\n\n\n\n<li><strong>Acceleration<\/strong>: a=mgM+ma = \\frac{mg}{M + m}a=M+mmg\u200b<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>This problem involves analyzing the motion of a coupled system: a cart of mass MMM connected to a hanging mass mmm via a pulley. Initially, the cart and mass are at rest, with the hanging mass elevated at height hhh. The gravitational potential energy of the hanging mass is the only mechanical energy present, as nothing is moving. The system\u2019s total energy is thus Ei=mghE_i = mghEi\u200b=mgh.<\/p>\n\n\n\n<p>When the mass is released and falls a distance ddd, it accelerates, pulling the cart horizontally. Both the cart and the mass move at the same speed vvv due to the inextensible rope. As the system moves, the gravitational potential energy of the hanging mass decreases, while kinetic energy increases for both masses. The final energy consists of the remaining potential energy of the hanging mass and the kinetic energy of both the cart and the mass.<\/p>\n\n\n\n<p>Applying conservation of energy gives an expression relating vvv, ggg, mmm, MMM, and ddd. Solving the equation yields the final velocity:v=2mgdM+mv = \\sqrt{\\frac{2mgd}{M + m}}v=M+m2mgd\u200b\u200b<\/p>\n\n\n\n<p>This shows that a greater mass mmm, a larger fall distance ddd, or a smaller cart mass MMM increases the final speed.<\/p>\n\n\n\n<p>Next, using the kinematic relation v2=2adv^2 = 2adv2=2ad, we isolate aaa, the system&#8217;s acceleration, and substitute the earlier expression for v2v^2v2. This leads to:a=mgM+ma = \\frac{mg}{M + m}a=M+mmg\u200b<\/p>\n\n\n\n<p>This is consistent with Newton&#8217;s second law and shows that acceleration depends on the net force (due to gravity on mass mmm) divided by the total mass of the system.<\/p>\n\n\n\n<p>This analysis shows how energy conservation can yield deep insight into motion, without requiring force diagrams or direct Newtonian analysis.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-731.jpeg\" alt=\"\" class=\"wp-image-230210\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The mass of the hanging mass is m and the mass of the cart is M. You can assume that the floor over which the mass hangs has been chosen as the zero point of gravitational potential energy. If the hanging mass is at a heightabove the floor, and has not yet been released, what [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-230209","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230209","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=230209"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230209\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230209"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230209"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230209"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}