{"id":88112,"date":"2025-06-01T07:01:46","date_gmt":"2025-06-01T07:01:46","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88112"},"modified":"2025-06-01T07:01:46","modified_gmt":"2025-06-01T07:01:46","slug":"a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/","title":{"rendered":"A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to"},"content":{"rendered":"

A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to rest on a rough horizontal surface after travelling a distance of 3 m. The magnitude of the frictional force is :<\/p>\n

[amp_mcq option1=”9 N” option2=”3 N” option3=”18 N” option4=”1 N” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nWe can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1\/2) * m * v_i^2 = (1\/2) * 2 kg * (3 m\/s)^2 = 9 J. Final KE = (1\/2) * m * v_f^2 = (1\/2) * 2 kg * (0 m\/s)^2 = 0 J. Change in KE = 0 – 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180\u00b0) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) \/ (-3 m) = 3 N. The magnitude of the frictional force is 3 N.
\n<\/section>\n
\n– Work-Energy Theorem: Net Work Done = Change in Kinetic Energy.
\n– Initial Kinetic Energy = (1\/2)mv_i^2.
\n– Final Kinetic Energy = (1\/2)mv_f^2.
\n– Work done by friction = – f * d.
\n<\/section>\n
\nAlternatively, one could calculate the deceleration using kinematics (v_f^2 = v_i^2 + 2ad) and then find the force using Newton’s second law (F=ma). 0^2 = 3^2 + 2 * a * 3 => 0 = 9 + 6a => 6a = -9 => a = -1.5 m\/s^2. The magnitude of acceleration is 1.5 m\/s^2. The net force in the direction of motion is the frictional force (acting opposite to initial motion). F_net = m * a = 2 kg * (-1.5 m\/s^2) = -3 N. The magnitude of the force is 3 N.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to rest on a rough horizontal surface after travelling a distance of 3 m. The magnitude of the frictional force is : [amp_mcq option1=”9 N” option2=”3 N” option3=”18 N” option4=”1 N” correct=”option2″] This question was previously asked in UPSC NDA-1 … <\/p>\n

Detailed SolutionA block of mass 2 kg, moving with the initial speed of 3 m\/s comes to<\/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":[1103,1129,1128],"class_list":["post-88112","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA block of mass 2 kg, moving with the initial speed of 3 m\/s comes to<\/title>\n<meta name=\"description\" content=\"We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1\/2) * m * v_i^2 = (1\/2) * 2 kg * (3 m\/s)^2 = 9 J. Final KE = (1\/2) * m * v_f^2 = (1\/2) * 2 kg * (0 m\/s)^2 = 0 J. Change in KE = 0 - 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180\u00b0) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) \/ (-3 m) = 3 N. The magnitude of the frictional force is 3 N. - Work-Energy Theorem: Net Work Done = Change in Kinetic Energy. - Initial Kinetic Energy = (1\/2)mv_i^2. - Final Kinetic Energy = (1\/2)mv_f^2. - Work done by friction = - f * d.\" \/>\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-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to\" \/>\n<meta property=\"og:description\" content=\"We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1\/2) * m * v_i^2 = (1\/2) * 2 kg * (3 m\/s)^2 = 9 J. Final KE = (1\/2) * m * v_f^2 = (1\/2) * 2 kg * (0 m\/s)^2 = 0 J. Change in KE = 0 - 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180\u00b0) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) \/ (-3 m) = 3 N. The magnitude of the frictional force is 3 N. - Work-Energy Theorem: Net Work Done = Change in Kinetic Energy. - Initial Kinetic Energy = (1\/2)mv_i^2. - Final Kinetic Energy = (1\/2)mv_f^2. - Work done by friction = - f * d.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:01:46+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":"A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to","description":"We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1\/2) * m * v_i^2 = (1\/2) * 2 kg * (3 m\/s)^2 = 9 J. Final KE = (1\/2) * m * v_f^2 = (1\/2) * 2 kg * (0 m\/s)^2 = 0 J. Change in KE = 0 - 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180\u00b0) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) \/ (-3 m) = 3 N. The magnitude of the frictional force is 3 N. - Work-Energy Theorem: Net Work Done = Change in Kinetic Energy. - Initial Kinetic Energy = (1\/2)mv_i^2. - Final Kinetic Energy = (1\/2)mv_f^2. - Work done by friction = - f * d.","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-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/","og_locale":"en_US","og_type":"article","og_title":"A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to","og_description":"We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1\/2) * m * v_i^2 = (1\/2) * 2 kg * (3 m\/s)^2 = 9 J. Final KE = (1\/2) * m * v_f^2 = (1\/2) * 2 kg * (0 m\/s)^2 = 0 J. Change in KE = 0 - 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180\u00b0) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) \/ (-3 m) = 3 N. The magnitude of the frictional force is 3 N. - Work-Energy Theorem: Net Work Done = Change in Kinetic Energy. - Initial Kinetic Energy = (1\/2)mv_i^2. - Final Kinetic Energy = (1\/2)mv_f^2. - Work done by friction = - f * d.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:01:46+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\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/","name":"A block of mass 2 kg, moving with the initial speed of 3 m\/s comes to","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:01:46+00:00","dateModified":"2025-06-01T07:01:46+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"We can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy. The block starts with an initial kinetic energy and comes to rest, meaning its final kinetic energy is zero. The change in kinetic energy is the final KE minus the initial KE. Initial KE = (1\/2) * m * v_i^2 = (1\/2) * 2 kg * (3 m\/s)^2 = 9 J. Final KE = (1\/2) * m * v_f^2 = (1\/2) * 2 kg * (0 m\/s)^2 = 0 J. Change in KE = 0 - 9 = -9 J. The only force doing work in the horizontal direction is the frictional force (f), opposing the motion. The work done by friction is W_friction = f * d * cos(180\u00b0) = -f * d, where d is the distance (3 m). According to the work-energy theorem, W_friction = Change in KE. So, -f * 3 m = -9 J. Solving for f, we get f = (-9 J) \/ (-3 m) = 3 N. The magnitude of the frictional force is 3 N. - Work-Energy Theorem: Net Work Done = Change in Kinetic Energy. - Initial Kinetic Energy = (1\/2)mv_i^2. - Final Kinetic Energy = (1\/2)mv_f^2. - Work done by friction = - f * d.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-block-of-mass-2-kg-moving-with-the-initial-speed-of-3-m-s-comes-to\/#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 block of mass 2 kg, moving with the initial speed of 3 m\/s comes to"}]},{"@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\/88112","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=88112"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88112\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}