{"id":88348,"date":"2025-06-01T07:08:30","date_gmt":"2025-06-01T07:08:30","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88348"},"modified":"2025-06-01T07:08:30","modified_gmt":"2025-06-01T07:08:30","slug":"the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/","title":{"rendered":"The force acting on a particle of mass m moving along the x-axis is gi"},"content":{"rendered":"

The force acting on a particle of mass m moving along the x-axis is given by F(x) = Ax2<\/sup> \u2013 Bx. Which one of the following is the potential energy of the particle ?<\/p>\n

[amp_mcq option1=”2Ax \u2013 B” option2=”$ -\\frac{x^2}{6} (2Ax \u2013 3B) $” option3=”$ Ax^3 \u2013 Bx^2 $” option4=”Zero” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2017<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -\\frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -\\int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 – Bx$, we integrate:
\n$U(x) = -\\int (Ax^2 – Bx) dx = – \\left( A \\int x^2 dx – B \\int x dx \\right) = – \\left( A \\frac{x^3}{3} – B \\frac{x^2}{2} \\right) + C$
\n$U(x) = -\\frac{A}{3}x^3 + \\frac{B}{2}x^2 + C$.
\nIf we set the integration constant $C=0$, the potential energy is $U(x) = -\\frac{A}{3}x^3 + \\frac{B}{2}x^2$. Let’s rewrite Option B: $ -\\frac{x^2}{6} (2Ax \u2013 3B) = -\\frac{2Ax^3}{6} + \\frac{3Bx^2}{6} = -\\frac{A}{3}x^3 + \\frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$.
\n<\/section>\n
\n– The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU\/dx$.
\n– To find $U(x)$ from $F(x)$, integrate: $U(x) = -\\int F(x) dx$.
\n<\/section>\n
\nThe integration constant $C$ represents the arbitrary choice of the zero point for potential energy. Different choices of $C$ result in different absolute values for potential energy, but the force (which depends on the *derivative* of potential energy, or the *difference* in potential energy between two points) remains the same. In multiple-choice questions involving potential energy derived from a force, the correct option often corresponds to setting the integration constant to zero or a value that makes the expression fit one of the choices.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The force acting on a particle of mass m moving along the x-axis is given by F(x) = Ax2 \u2013 Bx. Which one of the following is the potential energy of the particle ? [amp_mcq option1=”2Ax \u2013 B” option2=”$ -\\frac{x^2}{6} (2Ax \u2013 3B) $” option3=”$ Ax^3 \u2013 Bx^2 $” option4=”Zero” correct=”option2″] This question was previously … <\/p>\n

Detailed SolutionThe force acting on a particle of mass m moving along the x-axis is gi<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1101,1129,1128],"class_list":["post-88348","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1101","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe force acting on a particle of mass m moving along the x-axis is gi<\/title>\n<meta name=\"description\" content=\"The potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 - Bx$, we integrate: $U(x) = -int (Ax^2 - Bx) dx = - left( A int x^2 dx - B int x dx right) = - left( A frac{x^3}{3} - B frac{x^2}{2} right) + C$ $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2 + C$. If we set the integration constant $C=0$, the potential energy is $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2$. Let's rewrite Option B: $ -frac{x^2}{6} (2Ax \u2013 3B) = -frac{2Ax^3}{6} + frac{3Bx^2}{6} = -frac{A}{3}x^3 + frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$. - The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU\/dx$. - To find $U(x)$ from $F(x)$, integrate: $U(x) = -int F(x) dx$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The force acting on a particle of mass m moving along the x-axis is gi\" \/>\n<meta property=\"og:description\" content=\"The potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 - Bx$, we integrate: $U(x) = -int (Ax^2 - Bx) dx = - left( A int x^2 dx - B int x dx right) = - left( A frac{x^3}{3} - B frac{x^2}{2} right) + C$ $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2 + C$. If we set the integration constant $C=0$, the potential energy is $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2$. Let's rewrite Option B: $ -frac{x^2}{6} (2Ax \u2013 3B) = -frac{2Ax^3}{6} + frac{3Bx^2}{6} = -frac{A}{3}x^3 + frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$. - The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU\/dx$. - To find $U(x)$ from $F(x)$, integrate: $U(x) = -int F(x) dx$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:08:30+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"The force acting on a particle of mass m moving along the x-axis is gi","description":"The potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 - Bx$, we integrate: $U(x) = -int (Ax^2 - Bx) dx = - left( A int x^2 dx - B int x dx right) = - left( A frac{x^3}{3} - B frac{x^2}{2} right) + C$ $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2 + C$. If we set the integration constant $C=0$, the potential energy is $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2$. Let's rewrite Option B: $ -frac{x^2}{6} (2Ax \u2013 3B) = -frac{2Ax^3}{6} + frac{3Bx^2}{6} = -frac{A}{3}x^3 + frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$. - The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU\/dx$. - To find $U(x)$ from $F(x)$, integrate: $U(x) = -int F(x) dx$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/","og_locale":"en_US","og_type":"article","og_title":"The force acting on a particle of mass m moving along the x-axis is gi","og_description":"The potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 - Bx$, we integrate: $U(x) = -int (Ax^2 - Bx) dx = - left( A int x^2 dx - B int x dx right) = - left( A frac{x^3}{3} - B frac{x^2}{2} right) + C$ $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2 + C$. If we set the integration constant $C=0$, the potential energy is $U(x) = -frac{A}{3}x^3 + frac{B}{2}x^2$. Let's rewrite Option B: $ -frac{x^2}{6} (2Ax \u2013 3B) = -frac{2Ax^3}{6} + frac{3Bx^2}{6} = -frac{A}{3}x^3 + frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$. - The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU\/dx$. - To find $U(x)$ from $F(x)$, integrate: $U(x) = -int F(x) dx$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:08:30+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/","name":"The force acting on a particle of mass m moving along the x-axis is gi","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:08:30+00:00","dateModified":"2025-06-01T07:08:30+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -\\frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -\\int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 - Bx$, we integrate: $U(x) = -\\int (Ax^2 - Bx) dx = - \\left( A \\int x^2 dx - B \\int x dx \\right) = - \\left( A \\frac{x^3}{3} - B \\frac{x^2}{2} \\right) + C$ $U(x) = -\\frac{A}{3}x^3 + \\frac{B}{2}x^2 + C$. If we set the integration constant $C=0$, the potential energy is $U(x) = -\\frac{A}{3}x^3 + \\frac{B}{2}x^2$. Let's rewrite Option B: $ -\\frac{x^2}{6} (2Ax \u2013 3B) = -\\frac{2Ax^3}{6} + \\frac{3Bx^2}{6} = -\\frac{A}{3}x^3 + \\frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$. - The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU\/dx$. - To find $U(x)$ from $F(x)$, integrate: $U(x) = -\\int F(x) dx$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-force-acting-on-a-particle-of-mass-m-moving-along-the-x-axis-is-gi\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"The force acting on a particle of mass m moving along the x-axis is gi"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88348","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88348"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88348\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}