{"id":86812,"date":"2025-06-01T04:21:14","date_gmt":"2025-06-01T04:21:14","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=86812"},"modified":"2025-06-01T04:21:14","modified_gmt":"2025-06-01T04:21:14","slug":"a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/","title":{"rendered":"A stone of mass 1 kg and initially at rest is dropped from a tower of"},"content":{"rendered":"

A stone of mass 1 kg and initially at rest is dropped from a tower of height 40 m. When it reaches a height of 10 m from the ground level, what will be the values of its potential energy (PE) and kinetic energy (KE)? (Acceleration due to gravity is 10 m\/s\u00b2)<\/p>\n

[amp_mcq option1=”PE = 300 J, KE = 100 J” option2=”PE = 200 J, KE = 200 J” option3=”PE = 100 J, KE = 300 J” option4=”PE = 100 J, KE = 200 J” correct=”option3″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC Geoscientist – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance.
\nInitial state (at height 40 m, at rest):
\nInitial PE = mgh\u2081 = 1 kg * 10 m\/s\u00b2 * 40 m = 400 J
\nInitial KE = \u00bd mv\u2081\u00b2 = \u00bd * 1 kg * (0 m\/s)\u00b2 = 0 J
\nTotal Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J.
\nFinal state (at height 10 m):
\nFinal PE = mgh\u2082 = 1 kg * 10 m\/s\u00b2 * 10 m = 100 J.
\nSince Total Energy is conserved:
\nTotal Energy = Final PE + Final KE
\n400 J = 100 J + Final KE
\nFinal KE = 400 J – 100 J = 300 J.
\nSo, at a height of 10 m, PE = 100 J and KE = 300 J.
\n<\/section>\n
\nIn the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.
\n<\/section>\n
\nThe velocity of the stone at 10 m height can also be calculated from KE = \u00bd mv\u00b2, giving 300 J = \u00bd * 1 kg * v\u00b2, so v\u00b2 = 600 m\u00b2\/s\u00b2, and v = \u221a600 \u2248 24.5 m\/s. Alternatively, one could use kinematic equations (v\u00b2 = u\u00b2 + 2as) to find the velocity and then calculate KE. The energy conservation method is often simpler for problems involving height and speed changes.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A stone of mass 1 kg and initially at rest is dropped from a tower of height 40 m. When it reaches a height of 10 m from the ground level, what will be the values of its potential energy (PE) and kinetic energy (KE)? (Acceleration due to gravity is 10 m\/s\u00b2) [amp_mcq option1=”PE = … <\/p>\n

Detailed SolutionA stone of mass 1 kg and initially at rest is dropped from a tower of<\/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":[1110,1421,1128],"class_list":["post-86812","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1110","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nA stone of mass 1 kg and initially at rest is dropped from a tower of<\/title>\n<meta name=\"description\" content=\"The total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance. Initial state (at height 40 m, at rest): Initial PE = mgh\u2081 = 1 kg * 10 m\/s\u00b2 * 40 m = 400 J Initial KE = \u00bd mv\u2081\u00b2 = \u00bd * 1 kg * (0 m\/s)\u00b2 = 0 J Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J. Final state (at height 10 m): Final PE = mgh\u2082 = 1 kg * 10 m\/s\u00b2 * 10 m = 100 J. Since Total Energy is conserved: Total Energy = Final PE + Final KE 400 J = 100 J + Final KE Final KE = 400 J - 100 J = 300 J. So, at a height of 10 m, PE = 100 J and KE = 300 J. In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.\" \/>\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-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A stone of mass 1 kg and initially at rest is dropped from a tower of\" \/>\n<meta property=\"og:description\" content=\"The total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance. Initial state (at height 40 m, at rest): Initial PE = mgh\u2081 = 1 kg * 10 m\/s\u00b2 * 40 m = 400 J Initial KE = \u00bd mv\u2081\u00b2 = \u00bd * 1 kg * (0 m\/s)\u00b2 = 0 J Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J. Final state (at height 10 m): Final PE = mgh\u2082 = 1 kg * 10 m\/s\u00b2 * 10 m = 100 J. Since Total Energy is conserved: Total Energy = Final PE + Final KE 400 J = 100 J + Final KE Final KE = 400 J - 100 J = 300 J. So, at a height of 10 m, PE = 100 J and KE = 300 J. In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:21:14+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 stone of mass 1 kg and initially at rest is dropped from a tower of","description":"The total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance. Initial state (at height 40 m, at rest): Initial PE = mgh\u2081 = 1 kg * 10 m\/s\u00b2 * 40 m = 400 J Initial KE = \u00bd mv\u2081\u00b2 = \u00bd * 1 kg * (0 m\/s)\u00b2 = 0 J Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J. Final state (at height 10 m): Final PE = mgh\u2082 = 1 kg * 10 m\/s\u00b2 * 10 m = 100 J. Since Total Energy is conserved: Total Energy = Final PE + Final KE 400 J = 100 J + Final KE Final KE = 400 J - 100 J = 300 J. So, at a height of 10 m, PE = 100 J and KE = 300 J. In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.","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-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/","og_locale":"en_US","og_type":"article","og_title":"A stone of mass 1 kg and initially at rest is dropped from a tower of","og_description":"The total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance. Initial state (at height 40 m, at rest): Initial PE = mgh\u2081 = 1 kg * 10 m\/s\u00b2 * 40 m = 400 J Initial KE = \u00bd mv\u2081\u00b2 = \u00bd * 1 kg * (0 m\/s)\u00b2 = 0 J Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J. Final state (at height 10 m): Final PE = mgh\u2082 = 1 kg * 10 m\/s\u00b2 * 10 m = 100 J. Since Total Energy is conserved: Total Energy = Final PE + Final KE 400 J = 100 J + Final KE Final KE = 400 J - 100 J = 300 J. So, at a height of 10 m, PE = 100 J and KE = 300 J. In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:21:14+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-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/","name":"A stone of mass 1 kg and initially at rest is dropped from a tower of","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:21:14+00:00","dateModified":"2025-06-01T04:21:14+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance. Initial state (at height 40 m, at rest): Initial PE = mgh\u2081 = 1 kg * 10 m\/s\u00b2 * 40 m = 400 J Initial KE = \u00bd mv\u2081\u00b2 = \u00bd * 1 kg * (0 m\/s)\u00b2 = 0 J Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J. Final state (at height 10 m): Final PE = mgh\u2082 = 1 kg * 10 m\/s\u00b2 * 10 m = 100 J. Since Total Energy is conserved: Total Energy = Final PE + Final KE 400 J = 100 J + Final KE Final KE = 400 J - 100 J = 300 J. So, at a height of 10 m, PE = 100 J and KE = 300 J. In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-stone-of-mass-1-kg-and-initially-at-rest-is-dropped-from-a-tower-of\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"A stone of mass 1 kg and initially at rest is dropped from a tower of"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/86812","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=86812"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/86812\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=86812"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=86812"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=86812"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}