The elastic potential energy of a stretched spring is given by E = 50x2 . Where x is the displacement in meter and E is in joule, then the force constant of the spring is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
50 Nm
b
100 Nm-1
c
100 N/m2
d
100 Nm
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
U=12Kx2--(1), U=50x2--(2), compare equation (1) and (2) to find K