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Grade: 12

                        

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)

9 years ago

Answers : (2)

vikas askiitian expert
509 Points
							

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

9 years ago
Kushagra Madhukar
askIITians Faculty
605 Points
							
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/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
 
Thanks and regards,
Kushagra
 
3 months ago
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