A block is fastened to a horizontal spring. The block is pulled to a distance x = 10 cm from its equilibrium position (at x = 0 ) on a frictionless surface from rest. The kinetic energy of the block at x = 5 cm is 0.25 J. The spring constant of the spring is nearly (in Nm ⁻¹ )

Step-by-step solution

The amplitude of the motion is A = 10 cm = 0.1 m . The kinetic energy of a block in simple harmonic motion at a distance x from the equilibrium position is given by K = 1 2 k (A^2 - x^2) . Given K = 0.25 J at x = 5 cm = 0.05 m . Substituting the values: 0.25 = 1 2 k ((0.1)^2 - (0.05)^2) 0.5 = k (0.01 - 0.0025) 0.5 = k (0.0075) k = 0.5 0.0075 = 5000 75 = 200 3 66.67 Nm ⁻¹ The spring constant is nearly 67 Nm ⁻¹ . Answer: 67