Course ConNotes BookmarkE ?Notes Bookmark Evaluate EpCommunicate Print e) Info (

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Course ConNotes BookmarkE ?Notes Bookmark Evaluate EpCommunicate Print e) Info (8c8p29) A 2.00 kg block is placed against a spring on a frictionless 30° incline. The spring, whose spring constant is 19.8N/cm, is compressed 18.6 cm and then released. What is the elastic potential energy of the compressed spring? Submit Answer Tries 0/12 What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline? Submit AnswerTries 0/12 How far along the incline is the highest point from the release point? Submit AnswerTries 0/12 This discussion is closed. Send Feedback

Explanation / Answer

a) PE = 1/2 k x2

= 1/2 * 1980 * 0.1862

elastic potential energy = 34.25 J

b) change in gravitational potential energy = 34.25 J

c) h = PE / m g = 34.25 / (2.00 * 9.8) = 1.75 m

height = distance * sin 30

d = 1.75 / sin 30

= 3.49 m

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