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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$A$ force of $20\,dyne$ applied to the end of a spring increases its length by $1\,mm$. What will be the force constant of the spring?

$A$ block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has a force constant $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

$A$ block is attached to a spring as shown and very gradually lowered so that finally the spring expands by $d$. If the same block is attached to the spring and released suddenly,then the maximum expansion in the spring will be:

Find the acceleration of $3\,kg$ mass when the acceleration of $2\,kg$ mass is $2\,ms^{-2}$ as shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module