Can we apply hooke's law when a body of mass 1500g is hooked to its free end? what would happen in this case?​

The answer is c hope this

Answerthe force between the charges will be  f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes  two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges:     k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then   

F= (mass)(acceleration) = mam = 0.25 kgvi = 16 m/st = 2 svf = 0 m/s (since it was put to stop)a=(vf-vi)/ta=(0-16)/2a = 8 m/s^2 (decelerating)f = ma = (0.25 kg)(8 m/s^2)f = 2 nhope this !