|F| = (7 N/m)(4.00 times 10^-2 m) | - kx_max | = (7 N/m)(4.00 times 10^-2 m) = |

ID: 1496163 • Letter: #

Question

|F| = (7 N/m)(4.00 times 10^-2 m) | - kx_max | = (7 N/m)(4.00 times 10^-2 m) = |0.28| N (B) At what value of x is the force from the spring equal to 50% of the maximum force? Set | F | = | -kx | equal to 50% of the maximum value, without keeping track of the minus sign because the question asks about the magnitude of the force: | -kx | = 50/100=|kx_max| From this it is seen that the force is reduced to 50% of its maximum magnitude where |x| has 50% of its maximum value, or at x = + |3.5e-4 | cm.