Question

A 2.16 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 110 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 11.2 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Answer 1

0.15694 m

Step-by-step explanation:

m = Mass of block =
`2.16* 10^(-2)\ kg`

v = Velocity of block = 11.2 m/s

k = Spring constant = 110 N/m

Here the kinetic energy of the fall and spring are conserved

`(1)/(2)mv^2=(1)/(2)kA^2\\\Rightarrow mv^2=kA^2\\\Rightarrow A=\sqrt{(mv^2)/(k)}\\\Rightarrow A=\sqrt{(2.16* 10^(-2)* 11.2^2)/(110)}\\\Rightarrow A=0.15694\ m`

The amplitude of the resulting simple harmonic motion is 0.15694 m