# Initially ko 0 and uo kx2 mgyo where we neglect

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Unformatted text preview: + K = ~kX2 + mgyo. After the mass has left the spring, the potential energy is just the gravitational potential energy of the mass mgy, so mg(y - Yo) = ~kx2 - K. In terms of the distance, s, along the slope, y - Yo= (8 - So) sinO. The maximum distance occurs when the kinetic energy is zero, or (8 - SO)max = kx'l/2 mgsinO. Conservation of Mechanical til "skier starts down a frictionless 32 slope. After Yerticaldrop of 25 m, the slope temporarily y~ls out, then drops at 20 an additionaI 38 m 'callyhefore leveling out again. What is the r's speed on the two level stretches? ~tbn i,theslope is frictionless, mechanical energy is ed. Thus, I:1KAB == -AU AB, or == !mv~ Problem 23. A 12D-g arrow is shot vertically from a bow whose effective spring constant is 430 NJm. If the bow is drawn 71 cm before shooting the arrow, to what height does the arrow rise? -'YB)' is. Therefore, VB == V2(9.8 mJs2)(25 m) = Repeating for I:1KAG ==-D.UAC, we find 2 Vi<?-8 mJs )(63 m) = 35.1 m/s. (We assumed...
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## This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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