$\begin{array}{1 1}\large\frac{1}{18}\\\large\frac{1}{28}\\18\\28\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : $\large\frac{1}{18}$

From Hooke's law : $F=kx$ where $x$ is the distance the spring is stretched from its natural length,and $k$ is the (usually unknown) spring constant.

The spring's natural length is 30 cm,so it is stretched 12 cm from the natural length,we can use

$W=\int dW=\int F dx$

$2J=\int_0^{.12m}kx dx$

$\quad\;=\large\frac{k}{2}$$x^2\bigg|_0^{.12m}$

$\quad\;=\large\frac{k}{2}$$(0.0144m^2)$

$k=\large\frac{4Nm}{0.0144m^2}$

$\;\;\;=\large\frac{100}{9}$N/m

Once we have k,we can calculate the work done as follows

$W=\int dW$

$\;\;\;\;=\int\limits_0^{.1}F dx$

$\;\;\;\;=\int\limits_0^{.1} \large\frac{100}{9}$$xdx$

$\;\;\;\;=\large\frac{50}{9}$$x^2\bigg|_0^{.1}$

$\;\;\;\;=\large\frac{1}{18}$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...