{"id":195525,"date":"2025-02-28T11:55:23","date_gmt":"2025-02-28T11:55:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195525"},"modified":"2025-02-28T11:55:26","modified_gmt":"2025-02-28T11:55:26","slug":"two-blocks-connected-by-a-string-are-pulled-across-a-horizontal-surface-by-a-force-applied-to-one-of-the-blocks","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/two-blocks-connected-by-a-string-are-pulled-across-a-horizontal-surface-by-a-force-applied-to-one-of-the-blocks\/","title":{"rendered":"Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks"},"content":{"rendered":"\n

Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2.0m\/s2 to the right, what is the magnitude F of the applied force?<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

To determine the magnitude of the applied force ( F ) required to pull two blocks connected by a string across a horizontal surface with a given acceleration, we need to consider the forces acting on the system, including friction and the applied force’s components.<\/p>\n\n\n\n

Given Data:<\/strong><\/p>\n\n\n\n

    \n
  • Mass of block 1 (( m_1 )): 1.0 kg<\/li>\n\n\n\n
  • Mass of block 2 (( m_2 )): 3.0 kg<\/li>\n\n\n\n
  • Coefficient of kinetic friction (( \\mu_k )): 0.25<\/li>\n\n\n\n
  • Acceleration of the system (( a )): 2.0 m\/s\u00b2<\/li>\n\n\n\n
  • Angle of applied force (( \\theta )): 60\u00b0<\/li>\n<\/ul>\n\n\n\n

    Free-Body Diagrams:<\/strong><\/p>\n\n\n\n

      \n
    1. Block 1 (1.0 kg):<\/strong><\/li>\n<\/ol>\n\n\n\n
        \n
      • Forces:<\/em>\n
          \n
        • Tension (( T )) to the right<\/li>\n\n\n\n
        • Frictional force (( f_1 )) to the left<\/li>\n\n\n\n
        • Normal force (( N_1 )) upward<\/li>\n\n\n\n
        • Gravitational force (( W_1 = m_1 \\cdot g )) downward<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Block 2 (3.0 kg):<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • Forces:<\/em>\n
                \n
              • Applied force (( F )) at an angle ( \\theta = 60\u00b0 )<\/li>\n\n\n\n
              • Tension (( T )) to the left<\/li>\n\n\n\n
              • Frictional force (( f_2 )) to the left<\/li>\n\n\n\n
              • Normal force (( N_2 )) upward<\/li>\n\n\n\n
              • Gravitational force (( W_2 = m_2 \\cdot g )) downward<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

                Calculations:<\/strong><\/p>\n\n\n\n

                  \n
                1. Determine Normal Forces:<\/strong><\/li>\n<\/ol>\n\n\n\n
                    \n
                  • For block 1:
                    [
                    N_1 = W_1 = m_1 \\cdot g = 1.0\\, \\text{kg} \\times 9.8\\, \\text{m\/s}^2 = 9.8\\, \\text{N}
                    ]<\/li>\n\n\n\n
                  • For block 2, considering the vertical component of the applied force:
                    [
                    N_2 = W_2 – F \\cdot \\sin(\\theta) = m_2 \\cdot g – F \\cdot \\sin(60\u00b0)
                    ]
                    [
                    N_2 = 3.0\\, \\text{kg} \\times 9.8\\, \\text{m\/s}^2 – F \\cdot \\frac{\\sqrt{3}}{2} = 29.4\\, \\text{N} – 0.866F
                    ]<\/li>\n<\/ul>\n\n\n\n
                      \n
                    1. Calculate Frictional Forces:<\/strong><\/li>\n<\/ol>\n\n\n\n
                        \n
                      • Frictional force is given by ( f = \\mu_k \\cdot N ).<\/li>\n\n\n\n
                      • For block 1:
                        [
                        f_1 = \\mu_k \\cdot N_1 = 0.25 \\times 9.8\\, \\text{N} = 2.45\\, \\text{N}
                        ]<\/li>\n\n\n\n
                      • For block 2:
                        [
                        f_2 = \\mu_k \\cdot N_2 = 0.25 \\times (29.4\\, \\text{N} – 0.866F) = 7.35\\, \\text{N} – 0.2165F
                        ]<\/li>\n<\/ul>\n\n\n\n
                          \n
                        1. Apply Newton’s Second Law:<\/strong><\/li>\n<\/ol>\n\n\n\n
                            \n
                          • For block 1:
                            [
                            T – f_1 = m_1 \\cdot a
                            ]
                            [
                            T – 2.45\\, \\text{N} = 1.0\\, \\text{kg} \\times 2.0\\, \\text{m\/s}^2
                            ]
                            [
                            T = 4.45\\, \\text{N}
                            ]<\/li>\n\n\n\n
                          • For block 2:
                            [
                            F \\cdot \\cos(60\u00b0) – T – f_2 = m_2 \\cdot a
                            ]
                            [
                            0.5F – 4.45\\, \\text{N} – (7.35\\, \\text{N} – 0.2165F) = 3.0\\, \\text{kg} \\times 2.0\\, \\text{m\/s}^2
                            ]
                            [
                            0.5F – 4.45\\, \\text{N} – 7.35\\, \\text{N} + 0.2165F = 6.0\\, \\text{N}
                            ]
                            [
                            0.7165F – 11.8\\, \\text{N} = 6.0\\, \\text{N}
                            ]
                            [
                            0.7165F = 17.8\\, \\text{N}
                            ]
                            [
                            F = \\frac{17.8\\, \\text{N}}{0.7165} \\approx 24.85\\, \\text{N}
                            ]<\/li>\n<\/ul>\n\n\n\n

                            Conclusion:<\/strong>
                            The magnitude of the applied force ( F ) required to move the two blocks with an acceleration of 2.0 m\/s\u00b2, considering the given coefficient of kinetic friction, is approximately 24.85 N<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

                            Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2.0m\/s2 to the right, what is the magnitude F of the applied […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-195525","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195525","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=195525"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195525\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195525"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195525"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195525"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}