A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m.If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the amplitude of the oscillation is:
A)0.13 m
B)0.18 m
C)3.7 m
D)5.2 m
E)13 m

Question

Quizplus · Accepted Answer

The total mechanical energy of the system is given by $E = \frac{1}{2}kA^2 = \frac{1}{2}mv^2$. Solving for the amplitude, we get $A = \sqrt{\frac{mv^2}{k}}$. Plugging in the given values, we get $A = \sqrt{\frac{(0.25~kg)(3.0~m/s)^2}{200~N/m}} \approx 0.18~m$. Therefore, the amplitude of the oscillation is 0.18 m, which is answer choice B.

A 025-Kg Block Oscillates on the End of the Spring with with a Spring