A block of mass 2 kg is released from rest from a height of 0.15 m directly above an uncompressed vertical spring fixed to the ground. The spring has a spring constant of 200 N m ⁻¹ . The maximum compression of the spring is (Take g = 10 m s ⁻² )

Step-by-step solution

Let the maximum compression of the spring be x . Applying the principle of conservation of mechanical energy from the point of release to the point of maximum compression: Loss in gravitational potential energy = Gain in elastic potential energy mg(h + x) = 1 2 kx² Substituting the given values: 2 10 (0.15 + x) = 1 2 200 x² 20(0.15 + x) = 100x² 3 + 20x = 100x² 100x² - 20x - 3 = 0 This is a quadratic equation in x . Solving for x : 100x² - 30x + 10x - 3 = 0 10x(10x - 3) + 1(10x - 3) = 0 (10x + 1)(10x - 3) = 0 Since