A 12.0 cm-long spring has a force constant (k) of 400 N\/m. How much force is required to stretch the spring to a length of 14.0 cm?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To calculate the force required to stretch a spring, we use Hooke’s Law<\/strong>: F=k\u22c5xF = k \\cdot x<\/p>\n\n\n\n Where:<\/p>\n\n\n\n Given:<\/strong><\/p>\n\n\n\n F=400\u22c50.020=8.0 NF = 400 \\cdot 0.020 = \\boxed{8.0 \\text{ N}}<\/p>\n\n\n\n Springs obey Hooke\u2019s Law<\/strong>, which states that the force needed to extend or compress a spring by a certain distance is directly proportional to that distance, as long as the elastic limit is not exceeded. Mathematically, it is expressed as F=kxF = kx, where FF is the applied force, kk is the spring constant (a measure of the spring’s stiffness), and xx is the change in length from the spring\u2019s natural or unstressed length.<\/p>\n\n\n\n In this problem, the spring has a natural (unstretched) length of 12.0 cm, and we want to know how much force is needed to stretch it to 14.0 cm. The extension is the difference between these two lengths: 14.0\u2009cm\u221212.0\u2009cm=2.0\u2009cm=0.020\u2009m14.0\\, \\text{cm} – 12.0\\, \\text{cm} = 2.0\\, \\text{cm} = 0.020\\, \\text{m}<\/p>\n\n\n\n The spring constant is given as 400 N\/m, which means it takes 400 newtons of force to stretch the spring by 1 meter. Since we are only stretching it by 0.020 meters, we multiply: F=400\u00d70.020=8.0 NF = 400 \\times 0.020 = 8.0 \\text{ N}<\/p>\n\n\n\n This result shows that only 8.0 newtons<\/strong> of force are required to stretch the spring from 12.0 cm to 14.0 cm. The linear relationship between force and extension makes spring systems predictable and useful in various applications like measuring force (spring scales), cushioning (car suspensions), and mechanical clocks.<\/p>\n","protected":false},"excerpt":{"rendered":" A 12.0 cm-long spring has a force constant (k) of 400 N\/m. How much force is required to stretch the spring to a length of 14.0 cm? The correct answer and explanation is: Answer: To calculate the force required to stretch a spring, we use Hooke’s Law: F=k\u22c5xF = k \\cdot x Where: Given: F=400\u22c50.020=8.0 NF […]<\/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-18307","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18307","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=18307"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18307\/revisions"}],"predecessor-version":[{"id":18308,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18307\/revisions\/18308"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18307"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18307"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18307"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Answer:<\/strong><\/h3>\n\n\n\n
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\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n