Ask Questions, Get Answers

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

A spring whose natural length is 15 cm exerts a force of 45 N when stretched to a length of 20 cm.Find the work done in stretching the spring from a length of 20 cm to a length of 25 cm

$\begin{array}{1 1}3.375J\\2.375J\\4.375J\\3.415J\end{array} $

Can you answer this question?

1 Answer

0 votes
Answer : 3.375J
The change of length is 5cm =.05m.Since $F=kx$ we have
$\;\;\;=900$ N/m
The work done is
$W=900\int\limits_{0.05}^{.1} xdx$


answered Sep 2, 2014 by sreemathi.v
edited Sep 2, 2014 by balaji.thirumalai

Related questions

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