A particle can move along x-axis under the action of conservative force and its potential energy at any position is given by U=(x2−2x+16)J. Then the particle is in stable equilibrium at
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
x=1.5 m
b
x=0.75 m
c
x=1 m
d
x=0.5 m
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Force acting on the particle, F=−dUdx=(−2x+2)NAt equilibrium, F=0⇒−2x+2=0⇒x=1mNow, d2Udx2=−2. So potential energy is minimum at x = 1m.Hence at x = 1m, the particle is in stable equilibrium.
A particle can move along x-axis under the action of conservative force and its potential energy at any position is given by U=(x2−2x+16)J. Then the particle is in stable equilibrium at