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3 years ago

7

A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

ant is 125 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.8 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

1 answer:

avanturin [10]3 years ago

6 0

Answer:

0.17547 m

Explanation:

m = Mass of block = $3.3\times 10^{-2}\ kg$

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m$

The amplitude of the resulting simple harmonic motion is 0.17547 m

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<u>When m1<m2</u>

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