{"id":88178,"date":"2025-06-01T07:04:02","date_gmt":"2025-06-01T07:04:02","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88178"},"modified":"2025-06-01T07:04:02","modified_gmt":"2025-06-01T07:04:02","slug":"two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/","title":{"rendered":"Two forces, one of 3 newton and another of 4 newton are applied on a s"},"content":{"rendered":"

Two forces, one of 3 newton and another of 4 newton are applied on a standard 1 kg body, placed on a horizontal and frictionless surface, simultaneously along the x-axis and the y-axis, respectively, as shown below:
\nThe magnitude of the resultant acceleration is:<\/p>\n

[amp_mcq option1=”7 m\/s\u00b2” option2=”1 m\/s\u00b2” option3=”5 m\/s\u00b2” option4=”\u221a7 m\/s\u00b2” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2015<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is C, 5 m\/s\u00b2.
\n<\/section>\n
\nTwo forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $\\sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $\\sqrt{3^2 + 4^2} = \\sqrt{9 + 16} = \\sqrt{25} = 5$ N. According to Newton’s second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F\/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N \/ 1 kg = 5 m\/s\u00b2.
\n<\/section>\n
\nThe direction of the resultant force (and thus acceleration) would be at an angle to the x-axis, given by \u03b8 = atan(Fy\/Fx) = atan(4\/3). However, the question only asks for the magnitude of the acceleration. This problem illustrates how to find the resultant of perpendicular forces and apply Newton’s second law.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Two forces, one of 3 newton and another of 4 newton are applied on a standard 1 kg body, placed on a horizontal and frictionless surface, simultaneously along the x-axis and the y-axis, respectively, as shown below: The magnitude of the resultant acceleration is: [amp_mcq option1=”7 m\/s\u00b2” option2=”1 m\/s\u00b2” option3=”5 m\/s\u00b2” option4=”\u221a7 m\/s\u00b2” correct=”option3″] This … <\/p>\n

Detailed SolutionTwo forces, one of 3 newton and another of 4 newton are applied on a s<\/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":[1443,1129,1128],"class_list":["post-88178","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1443","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nTwo forces, one of 3 newton and another of 4 newton are applied on a s<\/title>\n<meta name=\"description\" content=\"The correct option is C, 5 m\/s\u00b2. Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$ N. According to Newton's second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F\/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N \/ 1 kg = 5 m\/s\u00b2.\" \/>\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\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two forces, one of 3 newton and another of 4 newton are applied on a s\" \/>\n<meta property=\"og:description\" content=\"The correct option is C, 5 m\/s\u00b2. Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$ N. According to Newton's second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F\/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N \/ 1 kg = 5 m\/s\u00b2.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:04:02+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":"Two forces, one of 3 newton and another of 4 newton are applied on a s","description":"The correct option is C, 5 m\/s\u00b2. Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$ N. According to Newton's second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F\/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N \/ 1 kg = 5 m\/s\u00b2.","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\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/","og_locale":"en_US","og_type":"article","og_title":"Two forces, one of 3 newton and another of 4 newton are applied on a s","og_description":"The correct option is C, 5 m\/s\u00b2. Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$ N. According to Newton's second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F\/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N \/ 1 kg = 5 m\/s\u00b2.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:04:02+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\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/","url":"https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/","name":"Two forces, one of 3 newton and another of 4 newton are applied on a s","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:04:02+00:00","dateModified":"2025-06-01T07:04:02+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct option is C, 5 m\/s\u00b2. Two forces are applied simultaneously and are perpendicular to each other (along the x and y axes). The resultant force is the vector sum of these forces. Since they are perpendicular, the magnitude of the resultant force (F_resultant) is found using the Pythagorean theorem: F_resultant = $\\sqrt{F_x^2 + F_y^2}$. Given F_x = 3 N and F_y = 4 N, F_resultant = $\\sqrt{3^2 + 4^2} = \\sqrt{9 + 16} = \\sqrt{25} = 5$ N. According to Newton's second law of motion, the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m) (a = F\/m). For a 1 kg body with a resultant force of 5 N, the magnitude of the acceleration is a = 5 N \/ 1 kg = 5 m\/s\u00b2.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-forces-one-of-3-newton-and-another-of-4-newton-are-applied-on-a-s\/#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":"Two forces, one of 3 newton and another of 4 newton are applied on a s"}]},{"@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\/88178","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=88178"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88178\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88178"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88178"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88178"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}