If the length of a $21 c m$ $21 cm$ spring increases to $45 c m$ $45 cm$ when a $2 k g$ $2 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $21 c m$ $21 cm$ spring increases to $45 c m$ $45 cm$ when a $2 k g$ $2 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2017-01-08T09:30:46

I got $k = 81.7 \frac{N}{m}$ $k=81.7N/m$

Explanation:

We can us Hooke's Law to relate the elastic force of a spring of elastic constant $k$ $k$ after being elongated of $x$ $x$ :
$F_{e l} = - k x$ $F_(el)=-kx$
Also, Newton's Second Law will help us to consider the forces involved through:
$SigmavecF=m veca$

let us see:
enter image source here
So at the equilibrium the elastic force and the weight ( $W=mg$ ) will give zero acceleration:
$-kx-mg=0$
But the elongation is directed downwards so we write, using numbers:
$-k(-0.24)-(2*9.8)=0$
So:
$k=(2*9.8)/0.24=81.7N/m$
I changed the lengths into meters because I am used to this but if you need it in N/cm use $24cm$ instead of $0.24m$ in the above.

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If the length of a $21 c m$ $21 cm$ spring increases to $45 c m$ $45 cm$ when a $2 k g$ $2 kg$ weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a 21 cm21cm21 cm spring increases to 45 cm45cm45 cm when a 2 kg2kg2 kg weight is hanging from it, what is the spring's constant?

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Explanation:

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If the length of a $21 c m$ $21 cm$ spring increases to $45 c m$ $45 cm$ when a $2 k g$ $2 kg$ weight is hanging from it, what is the spring's constant?