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

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Question

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

Vedclass · Accepted Answer

$5\beta - 4\alpha$

Vedclass · Answer

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

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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