logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A spring with a spring force constant $k=800\; N/m$ has a length of 5 cm. The work done in extending from 5 cm to 15 cm is

\[(a)\;16\;J \quad (b)\;8\;J \quad (c)\;32\;J \quad (d)\;24\;J\]

1 Answer

Answer: 8J
The work done by the spring force depends only on the end points. If the block is moved from an initial displacement $x_i$ to a final displacement $x_f$, the work done by the spring force $W_s$ is $\large\frac{1}{2} $$k (x_i^2 - x_f^2)$
$\Rightarrow W = \large\frac{1}{2} $$\times 800 \times (15^2-5^2)$$ = 8J$
answered Jul 17, 2013 by meena.p
edited Aug 25, 2014 by balaji.thirumalai
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X