During a race, you observe a 1500 kg car accelerate from an initial speed of 30m \/ s to a final speed of 60m \/ s Given this information, which of the following equations would be most useful for determining the work done accelerating the car?<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Correct Answer:<\/strong> Work = \u0394Kinetic Energy = (1\/2)mv\u00b2_final \u2212 (1\/2)mv\u00b2_initial<\/strong><\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n To determine the work done in accelerating a car, the most relevant concept is the work-energy principle<\/strong>. This principle states that the net work<\/strong> done on an object is equal to the change in its kinetic energy<\/strong>.<\/p>\n\n\n\n Kinetic energy is the energy an object possesses due to its motion and is given by the formula:<\/p>\n\n\n\n Kinetic Energy (KE) = (1\/2)mv\u00b2<\/strong>, In this case, the car has a mass of 1500 kg, an initial speed of 30 m\/s, and a final speed of 60 m\/s. Since the force and distance involved in the acceleration are not given, using the kinematic or force-based work formulas such as Work = Force \u00d7 Distance \u00d7 cos(\u03b8)<\/strong> would not be appropriate or useful. Instead, focusing on the change in kinetic energy<\/strong> provides a direct path to calculating the work done.<\/p>\n\n\n\n The change in kinetic energy<\/strong> (\u0394KE) is:<\/p>\n\n\n\n \u0394KE = (1\/2) * m * v\u00b2_final \u2212 (1\/2) * m * v\u00b2_initial This result means that 2,025,000 joules of work<\/strong> was done to increase the car\u2019s speed from 30 m\/s to 60 m\/s.<\/p>\n\n\n\n Therefore, the most useful equation is:<\/p>\n\n\n\n Work = (1\/2)mv\u00b2_final \u2212 (1\/2)mv\u00b2_initial<\/strong> During a race, you observe a 1500 kg car accelerate from an initial speed of 30m \/ s to a final speed of 60m \/ s Given this information, which of the following equations would be most useful for determining the work done accelerating the car? The correct answer and explanation is: Correct Answer:The most […]<\/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-30354","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30354","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=30354"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30354\/revisions"}],"predecessor-version":[{"id":30355,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30354\/revisions\/30355"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=30354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=30354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=30354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The most useful equation for determining the work done accelerating the car is:<\/p>\n\n\n\n
\n\n\n\n
where m<\/em> is the mass of the object and v<\/em> is its velocity.<\/p>\n\n\n\n
\u0394KE = (1\/2)(1500)(60)\u00b2 \u2212 (1\/2)(1500)(30)\u00b2
\u0394KE = (1\/2)(1500)(3600 \u2212 900)
\u0394KE = (750)(2700)
\u0394KE = 2,025,000 joules<\/p>\n\n\n\n
This equation directly links the car\u2019s mass and velocity to the amount of work performed.<\/p>\n","protected":false},"excerpt":{"rendered":"