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FrozenT [24]
3 years ago
12

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?
Physics
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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Given that,

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Distance can be calculated as,

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s = 81.55 m

For half height, that is, s = 40.77m

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2 * 9.81 * 40.77 = Vf² - 0²

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Vf = 28.28 m/s

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3 years ago
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Explanation:

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A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W
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Explanation:

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