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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
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
Dear student,Please find the solution to your problem. 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/2RKinetic energy at this instant is 0...Total energy = – GM/2R .................1Finally ball agian comes to rest , now ball is at a depth of R/2 from earth soIts gravitational energy is – 3GMm/2R[1 – r2/3R2]= – 3GMm/2R[11/12] (r = R/2)= – 11GMm/8Rpotential 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 energyTE = – 11GMm/8R + KR2/8 ................2total energy remains same so eq1 = eq 2KR2/8 – 11GMm/8R = – GMm/2RKR2/8 = 7GMm/8Ror, K = 7GMm/R3 Thanks and regards,Kushagra
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