{"id":88009,"date":"2025-06-01T06:59:34","date_gmt":"2025-06-01T06:59:34","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88009"},"modified":"2025-06-01T06:59:34","modified_gmt":"2025-06-01T06:59:34","slug":"one-block-of-2%c2%b70-kg-mass-is-placed-on-top-of-another-block-of-3%c2%b70-kg-m","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/one-block-of-2%c2%b70-kg-mass-is-placed-on-top-of-another-block-of-3%c2%b70-kg-m\/","title":{"rendered":"One block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg m"},"content":{"rendered":"

One block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg mass. The coefficient of static friction between the two blocks is 0\u00b72. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m\/s\u00b2, the maximum value of the frictional force is :<\/p>\n

[amp_mcq option1=”50 N” option2=”30 N” option3=”4 N” option4=”10 N” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nWhen the two blocks move together without slipping, the force responsible for the acceleration of the top block (mass 2.0 kg) is the static friction force exerted by the bottom block on the top block. The maximum value this static friction force can attain is given by the product of the coefficient of static friction and the normal force acting on the top block.
\n<\/section>\n
\n– The normal force on the top block is equal to its weight: N = m\u2081g.
\n– The maximum static friction force is given by $f_{s,max} = \\mu_s N$.
\n– For the blocks to move together without slipping, the actual static friction force must be less than or equal to $f_{s,max}$. The question asks for the maximum value of the frictional force in this scenario, which is the maximum static friction force that can exist between the blocks.
\n<\/section>\n
\nGiven: mass of top block m\u2081 = 2.0 kg, mass of bottom block m\u2082 = 3.0 kg, coefficient of static friction $\\mu_s = 0.2$, acceleration due to gravity g = 10 m\/s\u00b2.
\nNormal force on the top block (due to the bottom block) N = m\u2081g = 2.0 kg \u00d7 10 m\/s\u00b2 = 20 N.
\nMaximum static friction force $f_{s,max} = \\mu_s N = 0.2 \\times 20 N = 4 N$.
\nFor the blocks to move together, the actual friction force on the top block is $f = m_1 a$, where ‘a’ is the acceleration of the system. As the pulling force F on the bottom block increases, ‘a’ increases, and thus the required friction force ‘f’ increases. Slipping occurs when the required ‘f’ exceeds $f_{s,max}$. The maximum value the frictional force can reach while preventing slipping is $f_{s,max}$.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

One block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg mass. The coefficient of static friction between the two blocks is 0\u00b72. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m\/s\u00b2, … <\/p>\n

Detailed SolutionOne block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg m<\/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":[1105,1129,1128],"class_list":["post-88009","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nOne block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg m<\/title>\n<meta name=\"description\" content=\"When the two blocks move together without slipping, the force responsible for the acceleration of the top block (mass 2.0 kg) is the static friction force exerted by the bottom block on the top block. The maximum value this static friction force can attain is given by the product of the coefficient of static friction and the normal force acting on the top block. - The normal force on the top block is equal to its weight: N = m\u2081g. - The maximum static friction force is given by $f_{s,max} = mu_s N$. - For the blocks to move together without slipping, the actual static friction force must be less than or equal to $f_{s,max}$. The question asks for the maximum value of the frictional force in this scenario, which is the maximum static friction force that can exist between the blocks.\" \/>\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\/one-block-of-2\u00b70-kg-mass-is-placed-on-top-of-another-block-of-3\u00b70-kg-m\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"One block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg m\" \/>\n<meta property=\"og:description\" content=\"When the two blocks move together without slipping, the force responsible for the acceleration of the top block (mass 2.0 kg) is the static friction force exerted by the bottom block on the top block. The maximum value this static friction force can attain is given by the product of the coefficient of static friction and the normal force acting on the top block. - The normal force on the top block is equal to its weight: N = m\u2081g. - The maximum static friction force is given by $f_{s,max} = mu_s N$. - For the blocks to move together without slipping, the actual static friction force must be less than or equal to $f_{s,max}$. The question asks for the maximum value of the frictional force in this scenario, which is the maximum static friction force that can exist between the blocks.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/one-block-of-2\u00b70-kg-mass-is-placed-on-top-of-another-block-of-3\u00b70-kg-m\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:59:34+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":"One block of 2\u00b70 kg mass is placed on top of another block of 3\u00b70 kg m","description":"When the two blocks move together without slipping, the force responsible for the acceleration of the top block (mass 2.0 kg) is the static friction force exerted by the bottom block on the top block. 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