A body mass m is dropped from a certain height strikes a light vertical fixed spring of stiffness k.The height of its fall before touching the spring if the maximum compression of the spring is equal to 3mg/k is Ans-3mg/2k explain
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The change in potential energy between A and B = -mg
(h + x)
The change in K. E. between A and B = 0
The change in spring potential energy = (1/2) kx2
mg (h + x) + 0 + 1/2 kx2 = 0
h + x = k/2mg x2
h = k/2mg x2 - x
=(k/2mg) (3mg/k)2 - 3mg/k
h= 3mg/2k
The change in K. E. between A and B = 0
The change in spring potential energy = (1/2) kx2
mg (h + x) + 0 + 1/2 kx2 = 0
h + x = k/2mg x2
h = k/2mg x2 - x
=(k/2mg) (3mg/k)2 - 3mg/k
h= 3mg/2k
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