Question

A spring of force constant 800 n/m as an extension of 5cm. The work done in extending it from 5cm to 15cm ?

Answer 1

Answer:

8J

Explanation:

Potential Energy -

U_{f}-U_{i}= \int_{r_{i}}^{r_{f}}\vec{f}\cdot \vec{ds}

- wherein

U_{f}-final\: potential\: energy

U_{i}-initial \: potential\: energy

f-force

ds-small \: displacement

r_{i}-initial \: position

r_{f}-final\: position

Work done = change in energy stored in the spring.

=uf - ui

\omega=\frac{1}{2}kx_{2}^{2}-\frac{1}{2}kx_{i}^{2}

   =\frac{1}{2}\times800\times (15\times 10^{-2})^{2}-\frac{1}{2}\times 800\times(5\times10^{-2})^{2J}

   =400\times10^{-4}(225-25)=8J