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 ΔK = 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...
View
Full Document
 Fall '08
 Field
 Physics, Force, Kinetic Energy, Work, TA, horizontal frictionless surface, constant horizontal force, Tomoyuki Nakayama

Click to edit the document details