A spring is compressed 27.8 cm when a force of 76.5 N is applied. How fast will an 5.9 kg mass move if the spring is compressed 49.8 cm and released with the mass attached?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we use Hooke\u2019s Law<\/strong> and the Conservation of Energy<\/strong> principle.<\/p>\n\n\n\n Hooke\u2019s Law relates force and displacement for a spring: F=kxF = kx<\/p>\n\n\n\n Where:<\/p>\n\n\n\n Rearrange to solve for kk: k=Fx=76.50.278\u2248275.27\u2009N\/mk = \\frac{F}{x} = \\frac{76.5}{0.278} \\approx 275.27 \\, \\text{N\/m}<\/p>\n\n\n\n When the spring is compressed and then released, all the potential energy stored in the spring<\/strong> is converted into the kinetic energy<\/strong> of the attached mass: 12kx2=12mv2\\frac{1}{2}kx^2 = \\frac{1}{2}mv^2<\/p>\n\n\n\n Where:<\/p>\n\n\n\n 12(275.27)(0.498)2=12(5.9)v2\\frac{1}{2}(275.27)(0.498)^2 = \\frac{1}{2}(5.9)v^2<\/p>\n\n\n\n Simplify: 12(275.27)(0.248004)=12(5.9)v2\\frac{1}{2}(275.27)(0.248004) = \\frac{1}{2}(5.9)v^2 34.126\u22482.95v234.126 \\approx 2.95v^2 v2=34.1262.95\u224811.57v^2 = \\frac{34.126}{2.95} \\approx 11.57 v\u224811.57\u22483.40\u2009m\/sv \\approx \\sqrt{11.57} \\approx 3.40 \\, \\text{m\/s}<\/p>\n\n\n\n 3.40\u2009m\/s\\boxed{3.40 \\, \\text{m\/s}}<\/p>\n\n\n\n In this problem, we analyze how energy stored in a compressed spring is converted to motion. When a spring is compressed, it stores elastic potential energy<\/strong>. According to Hooke\u2019s Law<\/strong>, the force needed to compress a spring is directly proportional to the displacement: F=kxF = kx, where kk is the spring constant. Using the initial data \u2014 a force of 76.5 N causing a compression of 0.278 m \u2014 we calculate the spring constant as approximately 275.27 N\/m.<\/p>\n\n\n\n When the spring is compressed further (to 49.8 cm) and released with a 5.9 kg mass attached, that potential energy becomes kinetic energy<\/strong>. The formula for potential energy stored in a spring is: PE=12kx2PE = \\frac{1}{2}kx^2<\/p>\n\n\n\n This energy is transformed into kinetic energy of the moving mass: KE=12mv2KE = \\frac{1}{2}mv^2<\/p>\n\n\n\n Equating both (since energy is conserved), we find the final speed of the mass. Plugging in values and solving gives a final velocity of 3.40 m\/s<\/strong>.<\/p>\n\n\n\n This principle is central in physics and engineering, particularly in systems involving springs or elastic materials, from car suspensions to toys.<\/p>\n","protected":false},"excerpt":{"rendered":" A spring is compressed 27.8 cm when a force of 76.5 N is applied. How fast will an 5.9 kg mass move if the spring is compressed 49.8 cm and released with the mass attached? The correct answer and explanation is: To solve this problem, we use Hooke\u2019s Law and the Conservation of Energy principle. […]<\/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-19427","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19427","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=19427"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19427\/revisions"}],"predecessor-version":[{"id":19428,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19427\/revisions\/19428"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=19427"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=19427"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=19427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Find the spring constant kk<\/strong><\/h3>\n\n\n\n
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\n\n\n\nStep 2: Use energy conservation to find speed<\/strong><\/h3>\n\n\n\n
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\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\n\ud83d\udcd8 Explanation (300 words):<\/strong><\/h3>\n\n\n\n