## Essential University Physics: Volume 1 (3rd Edition)

a) $.367 \ m$ b) $567.6 \ N/m$
a) While there is a way to use complex math to arrive at the equation necessary to find this answer, we can also use proportions to obtain: $L = \frac{ x_1v_2^2- x_2v_1^2 }{v_2^2-v_1^2}$ $L = \frac{(.4)(12)^2 - (.6)(4.5)^2}{12^2-4.5^2}=.367 \ m$ b) We first find the value of $\mu$: $\mu = \frac{m}{L}=\frac{.34}{.367}=.925$ Thus, we can find k: $k = \frac{\mu v_1^2 }{\Delta x}$ $k = \frac{(.925) (4.5)^2 }{.033}=567.6 \ N/m$