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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
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