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3 years ago

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A 2.16 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

ant 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?

1 answer:

pochemuha3 years ago

3 0

Answer:

0.15694 m

Explanation:

m = Mass of block = $2.16\times 10^{-2}\ kg$

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

$\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow mv^2=kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{2.16\times 10^{-2}\times 11.2^2}{110}}\\\Rightarrow A=0.15694\ m$

The amplitude of the resulting simple harmonic motion is 0.15694 m

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