A block attached to a horizontal spring is pulled to x₀ = 10 cm and released from rest on a frictionless surface. When it reaches x = 5 cm, its kinetic energy is 0.25 J. Find the spring constant k in N m⁻¹, to the nearest integer.
Physics · Circular Motion
JEE Mains JEE Advanced

Question

A block attached to a horizontal spring is pulled to x₀ = 10 cm and released from rest on a frictionless surface. When it reaches x = 5 cm, its kinetic energy is 0.25 J. Find the spring constant k in N m⁻¹, to the nearest integer.

Physics · Circular Motion

G1 Physics · Accepted Answer

Initial total energy: E = ½kx₀² At x = x₀/2: E = ½k(x₀/2)² + 0.25 Therefore: ½kx₀² − ⅛kx₀² = 0.25 (3/8)kx₀² = 0.25 With x₀ = 0.10 m: k = 0.25 × 8 / [3 × (0.10)²] = 66.67 N m⁻¹ Nearest integer: 67. Note: the source page displays an inconsistent official value of 50; the shown energy-conservation calculation gives 67.