Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
+1 vote

It takes 100J of work to stretch a spring 0.5m from its equilibrium position.How much work is needed to stretch it an additional 0.75m?

$\begin{array}{1 1}525J\\625J\\725J\\325J\end{array} $

Can you answer this question?

1 Answer

0 votes
Answer : 525J
$100 =\int_0^{0.5} kxdx$
$\quad\;\;\;= \large\frac{1}{8}$$k$
$\Rightarrow k=800$
$W=\int\limits_{0.5}^{1.25} kxdx$
answered Sep 2, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App