If the length of a $38 c m$ $38 cm$ spring increases to $64 c m$ $64 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $38 c m$ $38 cm$ spring increases to $64 c m$ $64 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2018-03-18T14:09:01

We know.if by applying force $F$ $F$ we can cause $\partial x$ $del x$ amount of increase in the length of a spring,then they are related as $F = K \partial x$ $F=Kdel x$ (where, $K$ $K$ is the spring constant)

Given, $F = 4 \cdot 9.8 = 39.2 N$ $F=4*9.8=39.2N$ (as,here the weight of the object is the force which causes this extension)

and , $\partial x = \frac{64 - 38}{100} = 0.26 m$ $del x=(64-38)/100=0.26m$

so, $K = \frac{F}{\partial x} = \frac{39.2}{0.26} = 150.77 N m^{- 1}$ $K=F/(del x)=39.2/0.26=150.77 Nm^-1$

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If the length of a $38 c m$ $38 cm$ spring increases to $64 c m$ $64 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a 38cm38 cm spring increases to 64cm64 cm when a 4kg4 kg weight is hanging from it, what is the spring's constant?

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If the length of a $38 c m$ $38 cm$ spring increases to $64 c m$ $64 cm$ when a $4 k g$ $4 kg$ weight is hanging from it, what is the spring's constant?