ch08-p054

# ch08-p054 - d = 0.292 m being the only positive root(b We...

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54. (a) An appropriate picture (once friction is included) for this problem is Figure 8-3 in the textbook. We apply equation 8-31, Δ E th = f k d , and relate initial kinetic energy K i to the "resting" potential energy U r : K i + U i = f k d + K r + U r ¡ 20.0 J + 0 = f k d + 0 + 1 2 kd 2 where f k = 10.0 N and k = 400 N/m. We solve the equation for d using the quadratic formula or by using the polynomial solver on an appropriate calculator, with
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Unformatted text preview: d = 0.292 m being the only positive root. (b) We apply equation 8-31 again and relate U r to the "second" kinetic energy K s it has at the unstretched position. K r + U r = f k d + K s + U s ¡ 1 2 kd 2 = f k d + K s + 0 Using the result from part (a), this yields K s = 14.2 J....
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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