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A small ball of mass m is released at ht R above the surface of earth.If the maximum depth of the ball to which it goes is R/2 inside the earth through a marrow groove before coming to rest momentarily. The groovet, contains an ideal spring of spring constant k and natural length R find the value of k if R is the radius of earth and M mass of earth?
Ans: 7GMm/( R cube)
gravitational potential inside the earth is given by -3GM/2R [1-r2/3R2]
potential energy of ball when it was at a height of R above earth is-GMm/2R
kinetic energy at this instant is 0...
total energy = -GM/2R .................1
finally ball agian comes to rest , now ball is at a depth of R/2 from earth so
its gravitational energy is -3GMm/2R[1-r2/3R2]
=-3GMm/2R[11/12] (r=R/2)
=-11GMm/8R
potential energy stored in spring is KX2/2 = KR2/8 (compression is R/2)
total energy now is sum of potential energy of spring + gravitational potential energy
TE = -11GMm/8R + KR2/8 ................2
total energy remains same so eq1 = eq 2
KR2/8 -11GMm/8R = -GMm/2R
KR2 /8 = 7GMm/8R or
K = 7GMm/R3
approve my ans if u like
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