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3 years ago

7

A spring is hanging from the ceiling. When a 250 gram of mass is attached to the free end, the spring elongates by 5 cm. The spr

ing constant (in N/m) is:

1 answer:

lord [1]3 years ago

8 0

Answer:

k = 49 N/m

Explanation:

Given that,

Mass, m = 250 g = 0.25 kg

When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.

The force due to mass is balanced by its weight as follows :

mg=kx

$k=\dfrac{mg}{x}\\\\k=\dfrac{0.25\times 9.8}{0.05}\\\\k=49\ N/m$

So, the spring constant of the spring is 49 N/m.

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<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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