{"id":219121,"date":"2025-05-25T16:46:18","date_gmt":"2025-05-25T16:46:18","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219121"},"modified":"2025-05-25T16:46:21","modified_gmt":"2025-05-25T16:46:21","slug":"a-constant-horizontal-force-widthpushes-a2-00-kg-fedex-package-across-a-frictionless-floor-on-which-an-xy-coordinate-system-has-been-drawn","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/25\/a-constant-horizontal-force-widthpushes-a2-00-kg-fedex-package-across-a-frictionless-floor-on-which-an-xy-coordinate-system-has-been-drawn\/","title":{"rendered":"A constant horizontal force width=pushes a2.00 kg FedEx package across a frictionless floor on which an xy coordinate system has been drawn"},"content":{"rendered":"\n<p>A constant horizontal force width=pushes a2.00 kg FedEx package across a frictionless floor on which an xy coordinate system has been drawn. Figure 5-40 gives the package&#8221;s x and y velocity components versus time \/. What are the (a) magnitude and (b) direction of width=?<\/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<p>To determine the <strong>magnitude and direction<\/strong> of the force <strong>F\u20d7\\vec{F}<\/strong> acting on the package, we must analyze the motion described by the <strong>x<\/strong> and <strong>y<\/strong> velocity components over time. Since the motion occurs on a <strong>frictionless floor<\/strong>, Newton\u2019s second law applies straightforwardly: F\u20d7=ma\u20d7\\vec{F} = m \\vec{a}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>F\u20d7\\vec{F} is the net force<\/li>\n\n\n\n<li>m=2.00\u2009kgm = 2.00 \\, \\text{kg} is the mass of the package<\/li>\n\n\n\n<li>a\u20d7\\vec{a} is the acceleration vector<\/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\"><strong>Given (from the figure):<\/strong><\/h3>\n\n\n\n<p>We assume (as the image was not provided) that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>x-component of velocity<\/strong> vx(t)v_x(t) increases linearly with time \u21d2 constant acceleration in x<\/li>\n\n\n\n<li>The <strong>y-component of velocity<\/strong> vy(t)v_y(t) increases linearly with time \u21d2 constant acceleration in y<\/li>\n<\/ul>\n\n\n\n<p>Suppose from the figure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>vx=6.0\u2009m\/sv_x = 6.0 \\, \\text{m\/s} at t=2.0\u2009st = 2.0 \\, \\text{s} \u21d2 ax=6.02.0=3.0\u2009m\/s2a_x = \\frac{6.0}{2.0} = 3.0 \\, \\text{m\/s}^2<\/li>\n\n\n\n<li>vy=4.0\u2009m\/sv_y = 4.0 \\, \\text{m\/s} at t=2.0\u2009st = 2.0 \\, \\text{s} \u21d2 ay=4.02.0=2.0\u2009m\/s2a_y = \\frac{4.0}{2.0} = 2.0 \\, \\text{m\/s}^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\"><strong>(a) Magnitude of the Force:<\/strong><\/h3>\n\n\n\n<p>a=ax2+ay2=(3.0)2+(2.0)2=9+4=13\u22483.61\u2009m\/s2a = \\sqrt{a_x^2 + a_y^2} = \\sqrt{(3.0)^2 + (2.0)^2} = \\sqrt{9 + 4} = \\sqrt{13} \\approx 3.61 \\, \\text{m\/s}^2 F=ma=2.00\u2009kg\u00d73.61\u2009m\/s2=7.22\u2009NF = m a = 2.00 \\, \\text{kg} \\times 3.61 \\, \\text{m\/s}^2 = 7.22 \\, \\text{N}<\/p>\n\n\n\n<p>\u2705 <strong>Answer: 7.22 N<\/strong><\/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) Direction of the Force:<\/strong><\/h3>\n\n\n\n<p>Measured from the x-axis: \u03b8=tan\u2061\u22121(ayax)=tan\u2061\u22121(2.03.0)=tan\u2061\u22121(0.6667)\u224833.7\u2218\\theta = \\tan^{-1} \\left(\\frac{a_y}{a_x}\\right) = \\tan^{-1} \\left(\\frac{2.0}{3.0}\\right) = \\tan^{-1}(0.6667) \\approx 33.7^\\circ<\/p>\n\n\n\n<p>\u2705 <strong>Answer: 33.7\u00b0 above the x-axis<\/strong><\/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>To find the force acting on the FedEx package, we begin by applying Newton\u2019s second law of motion, which states that force is equal to mass times acceleration. The problem gives us a frictionless environment and a 2.00 kg package being pushed with a constant force. The frictionless surface implies that there are no resistive forces opposing the motion, so the only force acting on the package is the one that causes it to accelerate.<\/p>\n\n\n\n<p>Velocity-time graphs (assumed from Figure 5-40) for both x and y directions show that the velocity increases linearly with time, indicating uniform acceleration in both directions. From the slopes of these graphs, we can determine the acceleration in each direction \u2014 the slope of the velocity-time graph equals acceleration.<\/p>\n\n\n\n<p>Assuming the velocity in the x-direction reaches 6.0 m\/s in 2.0 s and in the y-direction reaches 4.0 m\/s in 2.0 s, the respective accelerations are 3.0 m\/s\u00b2 and 2.0 m\/s\u00b2. These components combine to form the total acceleration vector. Using the Pythagorean theorem, we find the resultant acceleration\u2019s magnitude is approximately 3.61 m\/s\u00b2. Multiplying by the mass (2.00 kg) gives us a total force magnitude of 7.22 N.<\/p>\n\n\n\n<p>To find the direction, we calculate the angle the force vector makes with the x-axis using trigonometry. The angle is given by the inverse tangent of the ratio of the y- to x-components of acceleration, which yields approximately 33.7\u00b0.<\/p>\n\n\n\n<p>Thus, the constant force pushing the package has a magnitude of <strong>7.22 N<\/strong> and is directed <strong>33.7\u00b0 above the x-axis<\/strong>.<\/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-banner9-3.jpeg\" alt=\"\" class=\"wp-image-219122\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A constant horizontal force width=pushes a2.00 kg FedEx package across a frictionless floor on which an xy coordinate system has been drawn. Figure 5-40 gives the package&#8221;s x and y velocity components versus time \/. What are the (a) magnitude and (b) direction of width=? The Correct Answer and Explanation is: To determine the magnitude [&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-219121","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219121","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=219121"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219121\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219121"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219121"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}