{"id":88008,"date":"2025-06-01T06:59:33","date_gmt":"2025-06-01T06:59:33","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88008"},"modified":"2025-06-01T06:59:33","modified_gmt":"2025-06-01T06:59:33","slug":"the-power-required-to-lift-a-mass-of-8%c2%b70-kg-up-a-vertical-distance-of","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-power-required-to-lift-a-mass-of-8%c2%b70-kg-up-a-vertical-distance-of\/","title":{"rendered":"The power required to lift a mass of 8\u00b70 kg up a vertical distance of"},"content":{"rendered":"

The power required to lift a mass of 8\u00b70 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m\/s\u00b2):<\/p>\n

[amp_mcq option1=”80 W” option2=”160 W” option3=”320 W” option4=”640 W” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe power required to lift an object is equal to the work done in lifting the object divided by the time taken. Work done in lifting is equal to the force required to lift it multiplied by the vertical distance lifted. The minimum force required to lift a mass vertically at a constant speed or to overcome gravity is equal to its weight.
\n<\/section>\n
\n– Weight (Force) = mass \u00d7 acceleration due to gravity (W = mg).
\n– Work done (W) = Force \u00d7 distance (W = Fd).
\n– Power (P) = Work done \/ Time (P = W\/t).
\n<\/section>\n
\nGiven: mass m = 8.0 kg, distance d = 4 m, time t = 2 s, acceleration due to gravity g = 10 m\/s\u00b2.
\nForce required to lift the mass = W = mg = 8.0 kg \u00d7 10 m\/s\u00b2 = 80 N.
\nWork done = Fd = 80 N \u00d7 4 m = 320 Joules.
\nPower = Work done \/ Time = 320 J \/ 2 s = 160 Watts.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The power required to lift a mass of 8\u00b70 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m\/s\u00b2): [amp_mcq option1=”80 W” option2=”160 W” option3=”320 W” option4=”640 W” correct=”option2″] This question was previously asked in UPSC NDA-1 – 2023 Download PDFAttempt Online The power required … <\/p>\n

Detailed SolutionThe power required to lift a mass of 8\u00b70 kg up a vertical distance 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":[1093],"tags":[1105,1129,1128],"class_list":["post-88008","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe power required to lift a mass of 8\u00b70 kg up a vertical distance of<\/title>\n<meta name=\"description\" content=\"The power required to lift an object is equal to the work done in lifting the object divided by the time taken. Work done in lifting is equal to the force required to lift it multiplied by the vertical distance lifted. The minimum force required to lift a mass vertically at a constant speed or to overcome gravity is equal to its weight. - Weight (Force) = mass \u00d7 acceleration due to gravity (W = mg). - Work done (W) = Force \u00d7 distance (W = Fd). - Power (P) = Work done \/ Time (P = W\/t).\" \/>\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\/the-power-required-to-lift-a-mass-of-8\u00b70-kg-up-a-vertical-distance-of\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The power required to lift a mass of 8\u00b70 kg up a vertical distance of\" \/>\n<meta property=\"og:description\" content=\"The power required to lift an object is equal to the work done in lifting the object divided by the time taken. Work done in lifting is equal to the force required to lift it multiplied by the vertical distance lifted. The minimum force required to lift a mass vertically at a constant speed or to overcome gravity is equal to its weight. - Weight (Force) = mass \u00d7 acceleration due to gravity (W = mg). - Work done (W) = Force \u00d7 distance (W = Fd). - Power (P) = Work done \/ Time (P = W\/t).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-power-required-to-lift-a-mass-of-8\u00b70-kg-up-a-vertical-distance-of\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:59:33+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":"The power required to lift a mass of 8\u00b70 kg up a vertical distance of","description":"The power required to lift an object is equal to the work done in lifting the object divided by the time taken. 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