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Unformatted text preview: Karen iF it has an inclination oF 33 . 9 ? Correct answer: 16884 . 9 J. Explanation: d h The escalator raised Karen a height oF h = d sin so the work done on Karen by the escalator is W = mgh = mgd sin = (51 . 4 kg) (9 . 8 m / s 2 ) (60 . 1 m) sin33 . 9 = 16884 . 9 J 003 10.0 points It takes 3 . 87 J oF work to stretch a Hookeslaw spring 8 . 5 cm From its unstressed length. How much the extra work is required to stretch it an additional 6 . 03 cm? Correct answer: 7 . 43848 J. Explanation: We begin by determining the value oF k From the energy equation solved For k , k = 2 U x 2 = 2(3 . 87 J) (0 . 085 m) 2 = 1071 . 28 J / m 2 We can now determine the energy at the additional displacement, U add = 1 2 k x 2 add = 1 2 1071 . 28 J / m 2 (0 . 085 m + 0 . 0603 m) 2 = 11 . 3085 J The extra work required is just the dierence in energy between the two displacements. W = U = U addU = 11 . 3085 J3 . 87 J = 7 . 43848 J...
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This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.
 Fall '09
 MIHALISIN
 Physics

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