{"id":246680,"date":"2025-07-07T10:31:59","date_gmt":"2025-07-07T10:31:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246680"},"modified":"2025-07-07T10:32:02","modified_gmt":"2025-07-07T10:32:02","slug":"at-an-instant-when-soccer-ball-is-in-contact-with-the-foot-of-a-player-kicking-it-the-horizontal-or-x-component-of-the-balls-acceleration-is-512-m-s","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/at-an-instant-when-soccer-ball-is-in-contact-with-the-foot-of-a-player-kicking-it-the-horizontal-or-x-component-of-the-balls-acceleration-is-512-m-s\/","title":{"rendered":"At an instant when soccer ball is in contact with the foot of a player kicking it, the horizontal or x component of the ball&#8217;s acceleration is 512 m\/s"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">At an instant when soccer ball is in contact with the foot of a player kicking it, the horizontal or x component of the ball&#8217;s acceleration is 512 m\/s? and the vertical or y component of its acceleration is 1400 m\/s? . The ball&#8217;s mass is 0.45 kg: What is the magnitude of the net force acting on the soccer ball at this instant?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">To find the magnitude of the net force acting on the soccer ball, we can use Newton\u2019s second law, which states that:F=maF = maF=ma<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>FFF is the net force,<\/li>\n\n\n\n<li>mmm is the mass of the object (in this case, the soccer ball), and<\/li>\n\n\n\n<li>aaa is the acceleration.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the total acceleration<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The components of the acceleration in the horizontal (x) and vertical (y) directions are given as:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>ax=512\u2009m\/s2a_x = 512 \\, \\text{m\/s}^2ax\u200b=512m\/s2<\/li>\n\n\n\n<li>ay=1400\u2009m\/s2a_y = 1400 \\, \\text{m\/s}^2ay\u200b=1400m\/s2<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">To find the total (or net) acceleration aaa, we combine these two components using the Pythagorean theorem, since the x and y directions are perpendicular to each other:a=ax2+ay2a = \\sqrt{a_x^2 + a_y^2}a=ax2\u200b+ay2\u200b\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substituting the given values:a=(512)2+(1400)2a = \\sqrt{(512)^2 + (1400)^2}a=(512)2+(1400)2\u200ba=262,144+1,960,000a = \\sqrt{262,144 + 1,960,000}a=262,144+1,960,000\u200ba=2,222,144a = \\sqrt{2,222,144}a=2,222,144\u200ba\u22481491.4\u2009m\/s2a \\approx 1491.4 \\, \\text{m\/s}^2a\u22481491.4m\/s2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Apply Newton&#8217;s second law<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Now that we have the total acceleration, we can apply Newton\u2019s second law to find the net force. The mass of the soccer ball is given as:m=0.45\u2009kgm = 0.45 \\, \\text{kg}m=0.45kg<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using the formula F=maF = maF=ma:F=(0.45\u2009kg)\u00d7(1491.4\u2009m\/s2)F = (0.45 \\, \\text{kg}) \\times (1491.4 \\, \\text{m\/s}^2)F=(0.45kg)\u00d7(1491.4m\/s2)F\u2248671.13\u2009NF \\approx 671.13 \\, \\text{N}F\u2248671.13N<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The magnitude of the net force acting on the soccer ball at this instant is approximately <strong>671.13 N<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The acceleration components in both the x and y directions were given. To find the total acceleration, we used the Pythagorean theorem because the accelerations are acting in perpendicular directions. After finding the net acceleration, we applied Newton\u2019s second law to calculate the net force, considering the ball\u2019s mass and the total acceleration it experiences. This method applies the principles of vector addition and Newton\u2019s second law of motion.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-915.jpeg\" alt=\"\" class=\"wp-image-246681\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>At an instant when soccer ball is in contact with the foot of a player kicking it, the horizontal or x component of the ball&#8217;s acceleration is 512 m\/s? and the vertical or y component of its acceleration is 1400 m\/s? . The ball&#8217;s mass is 0.45 kg: What is the magnitude of the net [&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-246680","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246680","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=246680"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246680\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246680"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246680"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246680"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}