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Unformatted text preview: K of the block as a function of position x . The block is initially at rest, thus the initial kinetic energy of the block is zero. Taking the origin at the relaxed position, the works done by the spring and the applied force are (1/2)kx 2 and Fx, respectively. Applying the workkinetic energy, we get K = W K = (1/2)kx 2 + Fx b) What is the magnitude of the force F ? We plug the values of graph at x = 1 m and x = 2 m in the equation in part a), and solve them for F. 8 =  (1/2)k 1 2 + F 1 16 =  k + 2F 0 = (1/2)k 2 2 + F 2 0 = 2k+2F From the second equation, we get k = F. Plugging this into the first equation, we have 16 =  F + 2F F = 16.0 N c) What is the spring constant k ? We already know that k = F from part b). Therefore, k = 16.0 N/m...
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This note was uploaded on 01/15/2012 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.
 Fall '08
 Field
 Physics, Work

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