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× [ =0.5×270×0.0046 =0.621 j
and Final PEf = 1/2 × K × = 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