{"id":87858,"date":"2025-06-01T06:55:39","date_gmt":"2025-06-01T06:55:39","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87858"},"modified":"2025-06-01T06:55:39","modified_gmt":"2025-06-01T06:55:39","slug":"a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/","title":{"rendered":"A mass M is dragged by a pulley on a horizontal plane by a force anti-"},"content":{"rendered":"

A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is<\/p>\n

[amp_mcq option1=”zero” option2=”positive” option3=”infinite” option4=”negative” correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2022<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nWork done (W) is calculated as the dot product of force (F) and displacement (d): W = F \u22c5 d = |F| |d| cos(\u03b8), where \u03b8 is the angle between the force vector and the displacement vector. The problem states that the force is “anti-parallel” to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180\u00b0)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative.
\n<\/section>\n
\nWork done is negative when the force and displacement are in opposite directions (anti-parallel).
\n<\/section>\n
\nWork done is positive when the force and displacement are in the same direction (parallel, \u03b8=0\u00b0, cos(0\u00b0)=1). Work done is zero when the force is perpendicular to the displacement (\u03b8=90\u00b0, cos(90\u00b0)=0).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is [amp_mcq option1=”zero” option2=”positive” option3=”infinite” option4=”negative” correct=”option4″] This question was previously asked in UPSC NDA-1 – 2022 Download PDFAttempt Online Work done (W) is calculated as the dot … <\/p>\n

Detailed SolutionA mass M is dragged by a pulley on a horizontal plane by a force anti-<\/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":[1093],"tags":[1108,1129,1128],"class_list":["post-87858","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1108","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA mass M is dragged by a pulley on a horizontal plane by a force anti-<\/title>\n<meta name=\"description\" content=\"Work done (W) is calculated as the dot product of force (F) and displacement (d): W = F \u22c5 d = |F| |d| cos(\u03b8), where \u03b8 is the angle between the force vector and the displacement vector. The problem states that the force is "anti-parallel" to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180\u00b0)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative. Work done is negative when the force and displacement are in opposite directions (anti-parallel).\" \/>\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\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A mass M is dragged by a pulley on a horizontal plane by a force anti-\" \/>\n<meta property=\"og:description\" content=\"Work done (W) is calculated as the dot product of force (F) and displacement (d): W = F \u22c5 d = |F| |d| cos(\u03b8), where \u03b8 is the angle between the force vector and the displacement vector. The problem states that the force is "anti-parallel" to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180\u00b0)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative. Work done is negative when the force and displacement are in opposite directions (anti-parallel).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:55:39+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=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A mass M is dragged by a pulley on a horizontal plane by a force anti-","description":"Work done (W) is calculated as the dot product of force (F) and displacement (d): W = F \u22c5 d = |F| |d| cos(\u03b8), where \u03b8 is the angle between the force vector and the displacement vector. The problem states that the force is \"anti-parallel\" to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180\u00b0)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative. Work done is negative when the force and displacement are in opposite directions (anti-parallel).","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\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/","og_locale":"en_US","og_type":"article","og_title":"A mass M is dragged by a pulley on a horizontal plane by a force anti-","og_description":"Work done (W) is calculated as the dot product of force (F) and displacement (d): W = F \u22c5 d = |F| |d| cos(\u03b8), where \u03b8 is the angle between the force vector and the displacement vector. The problem states that the force is \"anti-parallel\" to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180\u00b0)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative. Work done is negative when the force and displacement are in opposite directions (anti-parallel).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:55:39+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/","name":"A mass M is dragged by a pulley on a horizontal plane by a force anti-","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:55:39+00:00","dateModified":"2025-06-01T06:55:39+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Work done (W) is calculated as the dot product of force (F) and displacement (d): W = F \u22c5 d = |F| |d| cos(\u03b8), where \u03b8 is the angle between the force vector and the displacement vector. The problem states that the force is \"anti-parallel\" to its displacement. This means the force and displacement vectors are in opposite directions, so the angle between them is 180 degrees. The cosine of 180 degrees (cos(180\u00b0)) is -1. Therefore, the work done is W = |F| |d| (-1) = -|F| |d|. Since the magnitude of force and displacement are positive, the work done is negative. Work done is negative when the force and displacement are in opposite directions (anti-parallel).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-mass-m-is-dragged-by-a-pulley-on-a-horizontal-plane-by-a-force-anti\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"A mass M is dragged by a pulley on a horizontal plane by a force anti-"}]},{"@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\/87858","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=87858"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87858\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87858"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87858"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87858"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}