The length of a spring is alpha when a force | vedclass

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(A) Let the natural length of the spring be $\ell_{0}$ and its spring constant be $k$. According to Hooke's Law,$F = k(	ext{extension}) = k(\ell - \e

Vedclass · Accepted Answer

$5\beta - 4\alpha$

Vedclass · Answer

(A) Let the natural length of the spring be $\ell_{0}$ and its spring constant be $k$. According to Hooke's Law,$F = k(	ext{extension}) = k(\ell - \ell_{0})$.For the given cases:$4 = k(\alpha - \ell_{0}) \quad \dots(i)$$5 = k(\beta - \ell_{0}) \quad \dots(ii)$Let the length be $\gamma$ when the force is $9\,N$:$9 = k(\gamma - \ell_{0}) \quad \dots(iii)$Subtracting $(i)$ from $(ii)$:$5 - 4 = k(\beta - \ell_{0}) - k(\alpha - \ell_{0})$$1 = k(\beta - \alpha) \Rightarrow k = \frac{1}{\beta - \alpha}$From $(i)$,$\ell_{0} = \alpha - \frac{4}{k} = \alpha - 4(\beta - \alpha) = 5\alpha - 4\beta$.Substituting $k$ and $\ell_{0}$ into $(iii)$:$9 = \frac{1}{\beta - \alpha}(\gamma - (5\alpha - 4\beta))$$9(\beta - \alpha) = \gamma - 5\alpha + 4\beta$$\gamma = 9\beta - 9\alpha + 5\alpha - 4\beta = 5\beta - 4\alpha$.

The length of a spring is $\alpha$ when a force of $4\,N$ is applied on it and the length is $\beta$ when $5\,N$ force is applied. Then the length of the spring when $9\,N$ force is applied is

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