{"id":222086,"date":"2025-05-31T03:37:08","date_gmt":"2025-05-31T03:37:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222086"},"modified":"2025-05-31T03:37:10","modified_gmt":"2025-05-31T03:37:10","slug":"there-are-two-forces-on-the-2-00-kg-box-in-the-overhead-view-of-but-only-one-is-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/there-are-two-forces-on-the-2-00-kg-box-in-the-overhead-view-of-but-only-one-is-shown\/","title":{"rendered":"There are two forces on the 2.00 kg box in the overhead view of but only one is shown."},"content":{"rendered":"\n<p>There are two forces on the 2.00 kg box in the overhead view of but only one is shown. For F1 = 20.0N, a = 12.0 m\/s\u00b2 and ? = 30.0 degrees find the second force (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis.<\/p>\n\n\n\n<p>There are two forces on the 2.00 kg box in the overhead view of but only one is shown. For&nbsp;<em>F<\/em>1 = 20.0N,&nbsp;<em>a<\/em>&nbsp;= 12.0 m\/s\u00b2 and ? = 30.0 degrees find the second force (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the&nbsp;<em>x<\/em>&nbsp;axis.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of the box, m=2.00\u2009kgm = 2.00 \\, \\text{kg}<\/li>\n\n\n\n<li>Acceleration, a\u20d7=12.0\u2009m\/s2\\vec{a} = 12.0 \\, \\text{m\/s}^2<\/li>\n\n\n\n<li>One force, F\u20d71=20.0\u2009N\\vec{F}_1 = 20.0 \\, \\text{N} at \u03b8=30.0\u2218\\theta = 30.0^\\circ from the x-axis<\/li>\n\n\n\n<li>Need to find the second force F\u20d72\\vec{F}_2<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the Net Force Using Newton\u2019s Second Law<\/h3>\n\n\n\n<p>F\u20d7net=ma\u20d7=2.00\u00d712.0=24.0\u2009N\\vec{F}_{\\text{net}} = m \\vec{a} = 2.00 \\times 12.0 = 24.0 \\, \\text{N}<\/p>\n\n\n\n<p>This is the total net force acting on the box.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Resolve F\u20d71\\vec{F}_1 Into Components<\/h3>\n\n\n\n<p>F\u20d71=20.0cos\u2061(30\u2218)\u2009i^+20.0sin\u2061(30\u2218)\u2009j^=(17.32\u2009i^+10.0\u2009j^)\u2009N\\vec{F}_1 = 20.0 \\cos(30^\\circ) \\, \\hat{i} + 20.0 \\sin(30^\\circ) \\, \\hat{j} = (17.32 \\, \\hat{i} + 10.0 \\, \\hat{j}) \\, \\text{N}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Let F\u20d72=F2x\u2009i^+F2y\u2009j^\\vec{F}_2 = F_{2x} \\, \\hat{i} + F_{2y} \\, \\hat{j}<\/h3>\n\n\n\n<p>Then: F\u20d71+F\u20d72=F\u20d7net\u21d2F\u20d72=F\u20d7net\u2212F\u20d71\\vec{F}_1 + \\vec{F}_2 = \\vec{F}_{\\text{net}} \\Rightarrow \\vec{F}_2 = \\vec{F}_{\\text{net}} &#8211; \\vec{F}_1<\/p>\n\n\n\n<p>Let\u2019s assume net force is in some direction (unknown angle, but the magnitude is 24 N). Since only magnitude is given, let\u2019s assume it is entirely along the x-axis for simplicity (unless otherwise told, this is a standard assumption).<\/p>\n\n\n\n<p>So: F\u20d7net=24.0\u2009i^\\vec{F}_{\\text{net}} = 24.0 \\, \\hat{i}<\/p>\n\n\n\n<p>Now subtract: F\u20d72=(24.0\u2009i^)\u2212(17.32\u2009i^+10.0\u2009j^)=(6.68\u2009i^\u221210.0\u2009j^)\u2009N\\vec{F}_2 = (24.0 \\, \\hat{i}) &#8211; (17.32 \\, \\hat{i} + 10.0 \\, \\hat{j}) = (6.68 \\, \\hat{i} &#8211; 10.0 \\, \\hat{j}) \\, \\text{N}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(a) Answer in Unit Vector Notation:<\/strong><\/h3>\n\n\n\n<p>F\u20d72=6.68\u2009i^\u221210.0\u2009j^\u2009N\\boxed{\\vec{F}_2 = 6.68 \\, \\hat{i} &#8211; 10.0 \\, \\hat{j} \\, \\text{N}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(b) Magnitude of F\u20d72\\vec{F}_2:<\/strong><\/h3>\n\n\n\n<p>\u2223F\u20d72\u2223=(6.68)2+(\u221210.0)2=44.6+100=144.6\u224812.0\u2009N|\\vec{F}_2| = \\sqrt{(6.68)^2 + (-10.0)^2} = \\sqrt{44.6 + 100} = \\sqrt{144.6} \\approx \\boxed{12.0 \\, \\text{N}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>(c) Direction of F\u20d72\\vec{F}_2:<\/strong><\/h3>\n\n\n\n<p>\u03b8=tan\u2061\u22121(\u221210.06.68)=tan\u2061\u22121(\u22121.497)\u2248\u221256.3\u2218\\theta = \\tan^{-1}\\left(\\frac{-10.0}{6.68}\\right) = \\tan^{-1}(-1.497) \\approx -56.3^\\circ<\/p>\n\n\n\n<p>So, the angle is <strong>56.3\u00b0 below the +x axis<\/strong>, or: \u03b8=303.7\u2218&nbsp;counterclockwise&nbsp;from&nbsp;+x&nbsp;axis\\boxed{\\theta = 303.7^\\circ \\text{ counterclockwise from +x axis}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>This problem involves Newton\u2019s second law in vector form. We\u2019re told that a 2.00 kg box experiences two forces, one of which is given. The goal is to find the second force that causes a specific acceleration. First, we use Newton&#8217;s second law: F\u20d7net=ma\u20d7\\vec{F}_{\\text{net}} = m \\vec{a}. Since the mass is 2.00 kg and acceleration is 12.0 m\/s\u00b2, the net force must be 24.0 N.<\/p>\n\n\n\n<p>The given force F\u20d71\\vec{F}_1 has a magnitude of 20.0 N at 30\u00b0 to the x-axis, which we resolve into x and y components using trigonometry: F1x=20.0cos\u2061(30\u2218)=17.32\u2009N,F1y=20.0sin\u2061(30\u2218)=10.0\u2009NF_{1x} = 20.0 \\cos(30^\\circ) = 17.32\\, \\text{N}, \\quad F_{1y} = 20.0 \\sin(30^\\circ) = 10.0\\, \\text{N}<\/p>\n\n\n\n<p>To find the second force F\u20d72\\vec{F}_2, we subtract F\u20d71\\vec{F}_1 from the net force vector. Assuming the net force is along the x-axis (a common simplification unless a direction is specified), the net force is 24.0\u2009i^24.0 \\, \\hat{i}. Subtracting F\u20d71\\vec{F}_1 gives: F\u20d72=(24.0\u2009i^)\u2212(17.32\u2009i^+10.0\u2009j^)=6.68\u2009i^\u221210.0\u2009j^\\vec{F}_2 = (24.0 \\, \\hat{i}) &#8211; (17.32 \\, \\hat{i} + 10.0 \\, \\hat{j}) = 6.68 \\, \\hat{i} &#8211; 10.0 \\, \\hat{j}<\/p>\n\n\n\n<p>This result shows the second force acts partly forward and partly downward. We compute its magnitude with the Pythagorean theorem and direction using inverse tangent. The final direction is either as a negative angle from the x-axis or converted to a positive counterclockwise angle for standard notation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-164.jpeg\" alt=\"\" class=\"wp-image-222087\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>There are two forces on the 2.00 kg box in the overhead view of but only one is shown. For F1 = 20.0N, a = 12.0 m\/s\u00b2 and ? = 30.0 degrees find the second force (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of [&hellip;]<\/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-222086","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222086","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=222086"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222086\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222086"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222086"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}