Question

The work done in joules in increasing the extension of a spring of stiffness 10 N /cm from 4 cm to 6 cm is.

Answer 1

Answer:

1Joule

Explanation:

work done by spring = $\frac{1}{2}$K$(x_{2}^{2} - x_{1}^{2})$

where K is the stiffness of the spring = 10$\frac{N}{cm}$

           $x_{1}$ is initial position = 4cm

and      $x_{2}$ is final position = 6cm

since joule is in SI unit we need to convert all these into SI unit

1 cm = $\frac{1}{100}$m

so  K = $\frac{10}{\frac{1}{100} }$$\frac{N}{m}$ = 1000$\frac{N}{m}$

     $x_{1}$ = 0.04m

     $x_{2}$ = 0.06m

work done = $\frac{1}{2}$K$(x_{2}^{2} - x_{1}^{2})$

                  = $\frac{1}{2}$ * 1000* $[0.06^{2} - 0.04^{2}]$

                  = 1Joule