Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
a horizontal rod of mass m is hinged at 1end and the other end is conected to a vertical spring wth spring cnstant k..whn system is at eq. find the potential energy f the spring
Hello! Balancing the moments about the hinged end,you get mgl/2 = kxl ( since,weight at the centre of mass is mg and force at the other end is kx} from this,you get x=mg/2k now Potential energy of spring is 0.5kx^2,substituting the value of k,we get potential energy of spring = (mg)^2/8k THANKS ALL THE BEST PLZZZ APPROVE MY ANSWER IF YOU LIKE IT
Hello!
Balancing the moments about the hinged end,you get mgl/2 = kxl ( since,weight at the centre of mass is mg and force at the other end is kx}
from this,you get x=mg/2k
now Potential energy of spring is 0.5kx^2,substituting the value of k,we get
potential energy of spring = (mg)^2/8k
THANKS
ALL THE BEST
PLZZZ APPROVE MY ANSWER IF YOU LIKE IT
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -