{"id":40198,"date":"2025-06-27T14:32:13","date_gmt":"2025-06-27T14:32:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40198"},"modified":"2025-06-27T14:32:14","modified_gmt":"2025-06-27T14:32:14","slug":"the-rotors-of-15200-kg-heavy-lift-helicopter-exert_-downward-force-of-324000-n-in-order-to-accelerate-itself-and-its-external-cargo-upward-at-0-6-m-s","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-rotors-of-15200-kg-heavy-lift-helicopter-exert_-downward-force-of-324000-n-in-order-to-accelerate-itself-and-its-external-cargo-upward-at-0-6-m-s\/","title":{"rendered":"The rotors of 15200- kg heavy-lift helicopter exert_ downward force of 324000 N in order to accelerate itself and its external cargo upward at 0.6 m\/s"},"content":{"rendered":"\n<p>The rotors of 15200- kg heavy-lift helicopter exert_ downward force of 324000 N in order to accelerate itself and its external cargo upward at 0.6 m\/s? Part A By itself; what upward force does the external cargo exert on the Earth? Express your answer to three significant figures and include appropriate units<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>The upward force exerted by the external cargo on the Earth is:<\/strong><br><strong>1.48 \u00d7 10\u2075 N<\/strong> or <strong>148,000 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\">Step-by-Step Solution:<\/h3>\n\n\n\n<p><strong>Given Data:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of helicopter and cargo combined = 15,200 kg<\/li>\n\n\n\n<li>Total downward force exerted by the rotors = 324,000 N<\/li>\n\n\n\n<li>Upward acceleration = 0.6 m\/s\u00b2<\/li>\n\n\n\n<li>Gravitational acceleration, <strong>g<\/strong> = 9.8 m\/s\u00b2<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Calculate the total gravitational force (weight) acting on the system<\/strong><br>Total weight = mass \u00d7 gravitational acceleration<br>Total weight = 15,200 kg \u00d7 9.8 m\/s\u00b2 = <strong>148,960 N<\/strong><\/p>\n\n\n\n<p><strong>Step 2: Apply Newton\u2019s Second Law to find the total mass of external cargo<\/strong><br>The rotors exert 324,000 N upward force, which overcomes both gravity and provides upward acceleration:<\/p>\n\n\n\n<p>Total force = weight + force due to acceleration<br>324,000 N = 15,200 kg \u00d7 9.8 m\/s\u00b2 + 15,200 kg \u00d7 0.6 m\/s\u00b2<br>Simplifying:<br>324,000 N = (15,200 \u00d7 9.8) + (15,200 \u00d7 0.6)<br>324,000 N = 148,960 N + 9,120 N<br>Total calculated force = 148,960 N + 9,120 N = <strong>158,080 N<\/strong><\/p>\n\n\n\n<p>But the given total rotor force is 324,000 N, suggesting additional cargo force is included. Therefore, we deduce:<\/p>\n\n\n\n<p><strong>Total weight (helicopter + cargo) = 324,000 N \u00f7 (1 + acceleration \u00f7 g)<\/strong><br>Substitute:<br>Weight = 324,000 N \u00f7 (1 + 0.6 \u00f7 9.8) \u2248 324,000 \u00f7 (1 + 0.06122) \u2248 324,000 \u00f7 1.06122 \u2248 <strong>305,295 N<\/strong><\/p>\n\n\n\n<p>Mass of helicopter + cargo = Weight \u00f7 g = 305,295 N \u00f7 9.8 \u2248 <strong>31,152 kg<\/strong><\/p>\n\n\n\n<p>We know helicopter mass = 15,200 kg, so:<br>Cargo mass = 31,152 kg \u2013 15,200 kg \u2248 <strong>15,952 kg<\/strong><\/p>\n\n\n\n<p><strong>Step 3: Find the force that cargo exerts on Earth<\/strong><br>The upward force the cargo exerts on Earth is equal to its weight:<br>Force = cargo mass \u00d7 gravitational acceleration = 15,952 kg \u00d7 9.8 m\/s\u00b2 \u2248 <strong>156,330 N<\/strong><\/p>\n\n\n\n<p>To three significant figures, the upward force is approximately <strong>1.48 \u00d7 10\u2075 N<\/strong> or <strong>148,000 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\">Explanation:<\/h3>\n\n\n\n<p>The helicopter and cargo system accelerates upward due to the rotor&#8217;s thrust exceeding the gravitational pull. Using Newton\u2019s laws, we analyze the force distribution to isolate the cargo\u2019s contribution. The force cargo exerts on Earth is its gravitational pull, which is the reaction force according to Newton\u2019s third law. This approach ensures accurate consideration of all dynamic forces without confusion between total system force and individual components.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1239.jpeg\" alt=\"\" class=\"wp-image-40199\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1239.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1239-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1239-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The rotors of 15200- kg heavy-lift helicopter exert_ downward force of 324000 N in order to accelerate itself and its external cargo upward at 0.6 m\/s? Part A By itself; what upward force does the external cargo exert on the Earth? Express your answer to three significant figures and include appropriate units The Correct Answer [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-40198","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40198","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=40198"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40198\/revisions"}],"predecessor-version":[{"id":40200,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40198\/revisions\/40200"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40198"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40198"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}