A 620.0 g block of wood is firmly attached to a light horizontal spring with an elastic constant of 180.0 N\/m. When the system is compressed 5.0 cm and released, it elongates 2.3 cm beyond the equilibrium position before stopping and returning. What is the coefficient of kinetic friction of the floor? Write your answer to two decimal places.<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To find the coefficient of kinetic friction (\u03bc\u2096)<\/strong>, we can use the work-energy principle<\/strong>. The spring system starts compressed, moves through equilibrium, and stretches beyond it before stopping. The energy lost is due to friction<\/strong>, so:<\/p>\n\n\n\n When the spring is compressed 5.0 cm, the initial potential energy<\/strong> stored is: PE=12kx12=12(180.0)(0.050)2=0.225\u2009JPE = \\frac{1}{2} k x_1^2 = \\frac{1}{2} (180.0) (0.050)^2 = 0.225 \\, \\text{J}<\/p>\n\n\n\n At the maximum stretch (2.3 cm beyond equilibrium), the final potential energy<\/strong> is: PE=12kx22=12(180.0)(0.023)2=0.04761\u2009JPE = \\frac{1}{2} k x_2^2 = \\frac{1}{2} (180.0) (0.023)^2 = 0.04761 \\, \\text{J}<\/p>\n\n\n\n The work done by friction<\/strong> is equal to the initial PE \u2212 final PE<\/strong>: Wfriction=0.225\u22120.04761=0.17739\u2009JW_{\\text{friction}} = 0.225 – 0.04761 = 0.17739 \\, \\text{J}<\/p>\n\n\n\n fk=\u03bck\u22c5N=\u03bck\u22c5(mg)=\u03bck\u22c5(0.620\u22c59.8)=\u03bck\u22c56.076f_k = \\mu_k \\cdot N = \\mu_k \\cdot (m g) = \\mu_k \\cdot (0.620 \\cdot 9.8) = \\mu_k \\cdot 6.076<\/p>\n\n\n\n Wfriction=fk\u22c5d=\u03bck\u22c56.076\u22c50.073=0.443548\u03bckW_{\\text{friction}} = f_k \\cdot d = \\mu_k \\cdot 6.076 \\cdot 0.073 = 0.443548 \\mu_k<\/p>\n\n\n\n Set this equal to 0.17739 J: 0.443548\u03bck=0.177390.443548 \\mu_k = 0.17739<\/p>\n\n\n\n Solve for \u03bc\u2096<\/strong>: \u03bck=0.177390.443548\u22480.40\\mu_k = \\frac{0.17739}{0.443548} \\approx 0.40<\/p>\n\n\n\n Coefficient of kinetic friction \u03bc\u2096 = 0.40<\/strong><\/p>\n\n\n\n This problem uses the work-energy principle<\/strong> to find the coefficient of kinetic friction between a wooden block and a floor. When a spring is compressed and then released, it converts stored elastic potential energy into kinetic energy, which moves the block across the surface. However, because of friction<\/strong>, not all of that energy is conserved. Some is lost as thermal energy due to sliding.<\/p>\n\n\n\n Initially, when the spring is compressed, it holds a certain amount of potential energy<\/strong> based on the formula 12kx2\\frac{1}{2}kx^2. As the spring expands, that energy is transformed into motion. However, since the block doesn’t return all the way symmetrically, we can conclude that friction is opposing the motion<\/strong>.<\/p>\n\n\n\n By comparing the initial energy (compressed)<\/strong> and the final energy (after elongation)<\/strong>, we can calculate how much energy was lost due to friction<\/strong>. Then, using the formula for work (W=f\u22c5dW = f \\cdot d), we determine the frictional force. Since friction is also equal to the coefficient of kinetic friction multiplied by the normal force<\/strong> (which equals weight here), we solve for \u03bc\u2096<\/strong>.<\/p>\n\n\n\n This process elegantly demonstrates energy conservation in the presence of a non-conservative force<\/strong> like friction.<\/p>\n","protected":false},"excerpt":{"rendered":" A 620.0 g block of wood is firmly attached to a light horizontal spring with an elastic constant of 180.0 N\/m. When the system is compressed 5.0 cm and released, it elongates 2.3 cm beyond the equilibrium position before stopping and returning. What is the coefficient of kinetic friction of the floor? Write your answer […]<\/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-18495","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18495","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=18495"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18495\/revisions"}],"predecessor-version":[{"id":18496,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18495\/revisions\/18496"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18495"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18495"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18495"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Initial Energy (Spring Potential Energy)<\/h3>\n\n\n\n
\n\n\n\nStep 2: Final Energy<\/h3>\n\n\n\n
\n\n\n\nStep 3: Work Done by Friction<\/h3>\n\n\n\n
\n\n\n\nStep 4: Friction Force<\/h3>\n\n\n\n
\n\n\n\nStep 5: Work by Friction<\/h3>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/h3>\n\n\n\n
\n\n\n\n\ud83e\udde0 Explanation (300 words):<\/h3>\n\n\n\n