{"id":87167,"date":"2025-06-01T04:30:30","date_gmt":"2025-06-01T04:30:30","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87167"},"modified":"2025-06-01T04:30:30","modified_gmt":"2025-06-01T04:30:30","slug":"a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/","title":{"rendered":"A ball of mass 100 g falls freely from a height of h = 20 m and hits t"},"content":{"rendered":"

A ball of mass 100 g falls freely from a height of h = 20 m and hits the ground at a speed of 1\u00b74 \u221agh. Take g = 10 m s\u207b\u00b2. Which one among the following is the correct value of the work done on the ball by air friction ?<\/p>\n

[amp_mcq option1=”0\u00b74 J” option2=”0\u00b75 J” option3=”0\u00b73 J” option4=”1 J” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction.
\n<\/section>\n
\nWork done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m\/s\u00b2)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = \u00bd m v_f\u00b2, where v_f = 1.4\u221agh. With g=10 m\/s\u00b2 and h=20m, v_f = 1.4\u221a(10*20) = 1.4\u221a200. So, v_f\u00b2 = (1.4)\u00b2 * 200 = 1.96 * 200 = 392 m\u00b2\/s\u00b2. KE_f = \u00bd (0.1 kg)(392 m\u00b2\/s\u00b2) = 0.05 * 392 = 19.6 J.
\n<\/section>\n
\nAccording to the Work-Energy Theorem, W_total = W_g + W_air = \u0394KE = KE_f – KE_i. 20 J + W_air = 19.6 J – 0 J. W_air = 19.6 J – 20 J = -0.4 J. The question asks for the value, and options are positive, implying magnitude. The magnitude of the work done by air friction is |-0.4 J| = 0.4 J.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A ball of mass 100 g falls freely from a height of h = 20 m and hits the ground at a speed of 1\u00b74 \u221agh. Take g = 10 m s\u207b\u00b2. Which one among the following is the correct value of the work done on the ball by air friction ? [amp_mcq option1=”0\u00b74 J” … <\/p>\n

Detailed SolutionA ball of mass 100 g falls freely from a height of h = 20 m and hits t<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1091],"tags":[1103,1129,1128],"class_list":["post-87167","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA ball of mass 100 g falls freely from a height of h = 20 m and hits t<\/title>\n<meta name=\"description\" content=\"The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction. Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m\/s\u00b2)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = \u00bd m v_f\u00b2, where v_f = 1.4\u221agh. With g=10 m\/s\u00b2 and h=20m, v_f = 1.4\u221a(10*20) = 1.4\u221a200. So, v_f\u00b2 = (1.4)\u00b2 * 200 = 1.96 * 200 = 392 m\u00b2\/s\u00b2. KE_f = \u00bd (0.1 kg)(392 m\u00b2\/s\u00b2) = 0.05 * 392 = 19.6 J.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A ball of mass 100 g falls freely from a height of h = 20 m and hits t\" \/>\n<meta property=\"og:description\" content=\"The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction. Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m\/s\u00b2)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = \u00bd m v_f\u00b2, where v_f = 1.4\u221agh. With g=10 m\/s\u00b2 and h=20m, v_f = 1.4\u221a(10*20) = 1.4\u221a200. So, v_f\u00b2 = (1.4)\u00b2 * 200 = 1.96 * 200 = 392 m\u00b2\/s\u00b2. KE_f = \u00bd (0.1 kg)(392 m\u00b2\/s\u00b2) = 0.05 * 392 = 19.6 J.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:30:30+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A ball of mass 100 g falls freely from a height of h = 20 m and hits t","description":"The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction. Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m\/s\u00b2)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = \u00bd m v_f\u00b2, where v_f = 1.4\u221agh. With g=10 m\/s\u00b2 and h=20m, v_f = 1.4\u221a(10*20) = 1.4\u221a200. So, v_f\u00b2 = (1.4)\u00b2 * 200 = 1.96 * 200 = 392 m\u00b2\/s\u00b2. KE_f = \u00bd (0.1 kg)(392 m\u00b2\/s\u00b2) = 0.05 * 392 = 19.6 J.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/","og_locale":"en_US","og_type":"article","og_title":"A ball of mass 100 g falls freely from a height of h = 20 m and hits t","og_description":"The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction. Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m\/s\u00b2)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = \u00bd m v_f\u00b2, where v_f = 1.4\u221agh. With g=10 m\/s\u00b2 and h=20m, v_f = 1.4\u221a(10*20) = 1.4\u221a200. So, v_f\u00b2 = (1.4)\u00b2 * 200 = 1.96 * 200 = 392 m\u00b2\/s\u00b2. KE_f = \u00bd (0.1 kg)(392 m\u00b2\/s\u00b2) = 0.05 * 392 = 19.6 J.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:30:30+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-mass-100-g-falls-freely-from-a-height-of-h-20-m-and-hits-t\/","name":"A ball of mass 100 g falls freely from a height of h = 20 m and hits t","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:30:30+00:00","dateModified":"2025-06-01T04:30:30+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction. Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m\/s\u00b2)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = \u00bd m v_f\u00b2, where v_f = 1.4\u221agh. With g=10 m\/s\u00b2 and h=20m, v_f = 1.4\u221a(10*20) = 1.4\u221a200. So, v_f\u00b2 = (1.4)\u00b2 * 200 = 1.96 * 200 = 392 m\u00b2\/s\u00b2. 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