{"id":238898,"date":"2025-06-18T13:09:49","date_gmt":"2025-06-18T13:09:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=238898"},"modified":"2025-06-18T13:09:51","modified_gmt":"2025-06-18T13:09:51","slug":"the-acceleration-due-to-gravity-at-sea-level-is-g-32-2-ft-s2-and-decreases-by-0-003-ft-s2-per-1000-ft-of-elevation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/18\/the-acceleration-due-to-gravity-at-sea-level-is-g-32-2-ft-s2-and-decreases-by-0-003-ft-s2-per-1000-ft-of-elevation\/","title":{"rendered":"The acceleration due to gravity at sea level is g = 32.2 ft\/s^2 and decreases by 0.003 ft\/s^2 per 1000 ft of elevation."},"content":{"rendered":"\n

The acceleration due to gravity at sea level is g = 32.2 ft\/s^2 and decreases by 0.003 ft\/s^2 per 1000 ft of elevation. Find the weight (in lbf) of a 165 lbm man atop the 8849 meters Mt. Everest relative to this point? Full solution.<\/p>\n\n\n\n

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

To find the weight (in pounds-force, lbf<\/strong>) of a 165 lbm (pound-mass) man atop Mt. Everest<\/strong>, we account for the reduction in gravitational acceleration with altitude.<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Sea-level gravity, g0=32.2\u2009ft\/s2g_0 = 32.2 \\, \\text{ft\/s}^2g0\u200b=32.2ft\/s2<\/li>\n\n\n\n
  • Gravity decreases by 0.003 ft\/s\u00b2 per 1000 ft<\/li>\n\n\n\n
  • Elevation of Mt. Everest: 8849 meters<\/li>\n\n\n\n
  • 1 meter = 3.28084 feet<\/li>\n\n\n\n
  • Mass of man: m=165\u2009lbmm = 165 \\, \\text{lbm}m=165lbm<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 1: Convert elevation to feet<\/strong><\/h3>\n\n\n\n

    8849\u2009m\u00d73.28084\u2009ftm=29,032\u2009ft (approx.)8849 \\, \\text{m} \\times 3.28084 \\, \\frac{\\text{ft}}{\\text{m}} = 29,032 \\, \\text{ft (approx.)}8849m\u00d73.28084mft\u200b=29,032ft (approx.)<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Calculate reduction in gravitational acceleration<\/strong><\/h3>\n\n\n\n

    Reduction=(29,0321000)\u00d70.003=29.032\u00d70.003=0.0871\u2009ft\/s2\\text{Reduction} = \\left(\\frac{29,032}{1000}\\right) \\times 0.003 = 29.032 \\times 0.003 = 0.0871 \\, \\text{ft\/s}^2Reduction=(100029,032\u200b)\u00d70.003=29.032\u00d70.003=0.0871ft\/s2<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Gravity at the summit<\/strong><\/h3>\n\n\n\n

    gEverest=32.2\u22120.0871=32.1129\u2009ft\/s2g_{\\text{Everest}} = 32.2 – 0.0871 = 32.1129 \\, \\text{ft\/s}^2gEverest\u200b=32.2\u22120.0871=32.1129ft\/s2<\/p>\n\n\n\n

    \n\n\n\n

    Step 4: Weight on Mt. Everest<\/strong><\/h3>\n\n\n\n

    Weight in pounds-force is given by:W=m\u00d7gEverest\/g0\u00d7g0=m\u00d7(gEverestg0)\u00d7g0W = m \\times g_{\\text{Everest}} \/ g_0 \\times g_0 = m \\times \\left(\\frac{g_{\\text{Everest}}}{g_0}\\right) \\times g_0W=m\u00d7gEverest\u200b\/g0\u200b\u00d7g0\u200b=m\u00d7(g0\u200bgEverest\u200b\u200b)\u00d7g0\u200b<\/p>\n\n\n\n

    But since lbf is defined as lbm \u00d7 (local g) \/ standard g, we can simplify as:W=m\u00d7(gEverest32.2)=165\u00d7(32.112932.2)W = m \\times \\left(\\frac{g_{\\text{Everest}}}{32.2}\\right) = 165 \\times \\left(\\frac{32.1129}{32.2}\\right)W=m\u00d7(32.2gEverest\u200b\u200b)=165\u00d7(32.232.1129\u200b)W\u2248165\u00d70.9969=164.5\u2009lbfW \\approx 165 \\times 0.9969 = 164.5 \\, \\text{lbf}W\u2248165\u00d70.9969=164.5lbf<\/p>\n\n\n\n

    \n\n\n\n

    Final Answer:<\/strong><\/h3>\n\n\n\n

    The man\u2019s weight atop Mt. Everest is approximately 164.5 lbf.<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Explanation:<\/strong><\/h3>\n\n\n\n

    Weight is a measure of force, which depends on both mass and the local acceleration due to gravity. While a person\u2019s mass remains constant regardless of location, their weight can vary depending on the gravitational pull. At higher elevations like Mt. Everest, gravity weakens slightly because the distance from Earth’s center increases.<\/p>\n\n\n\n

    Here, we calculated how gravity changes with altitude based on the given rate of decrease. After converting the mountain’s height from meters to feet, we determined the total reduction in gravity. By subtracting this from the standard gravity at sea level, we obtained the adjusted gravitational value at the summit. Finally, multiplying this value by the man\u2019s mass in pound-mass and adjusting proportionally, we found his weight in pound-force.<\/p>\n\n\n\n

    This problem shows how physics must account for environmental conditions even when dealing with simple quantities like weight.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The acceleration due to gravity at sea level is g = 32.2 ft\/s^2 and decreases by 0.003 ft\/s^2 per 1000 ft of elevation. Find the weight (in lbf) of a 165 lbm man atop the 8849 meters Mt. Everest relative to this point? Full solution. The Correct Answer and Explanation is: To find the weight […]<\/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-238898","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238898","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=238898"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238898\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238898"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238898"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238898"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}