Question

A pen contains a spring with a spring constant of 270 N/m. When the tip of the pen is in its retracted position, the spring is compressed 4.60 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.70 mm. How much work is done by the spring force to ready the pen for writing?

Answer 1

Answer:

-0.9045 j

Explanation:

Here K= 270 N/m

distance 1 = 4.60 mm=0.0046 m

when pen is compressed ;

distance 2 = 6.70 mm + distance1 = 0.0067 m+0.0046 m= 0.0113 m

The pen store in itself two types of PE

Which is Initial PEi = 1/2 ×K× [$distance 1^{2}$ =0.5×270×0.0046 =0.621 j

and Final PEf = 1/2 × K × $distance 2^{2}$ = 0.5×270×0.0113=1.5255 j

When the pen gets ready - it goes ahead distance d2 and is then pressed back distance d1 for writing.

Total PE when ready  = 0.621-1.5255= - 0.9045 j

so, work =  -0.9045 j