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A block with a mass of 0.28 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.0 N on the block. When the block is released, it oscillates with a frequency of 1.2 Hz. How far was the block pulled back before being released?

Respuesta :

Answer:

Explanation:

For spring

[tex]n=\sqrt{\frac{k}{m} }[/tex]

where n is frequency of oscillation and k is force constant and m is mass

Putting the values

[tex]1.2=\sqrt{\frac{k}{.28} }[/tex]

k = .4032 N/m

F= k x

where F is force , k is force constant and x is extension

Putting the given values

1 = .4032 x

x = 2.48 m

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