A 4 kg block on smooth horizontal ground is attached to a spring of stiffness 100 N cm⁻¹. Spring free end is pulled to the right at 2 m s⁻¹; at t = 0 spring is unstretched and block moves left at 1 m s⁻¹. Find maximum spring extension (cm). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A 4 kg block on smooth horizontal ground is attached to a spring of stiffness 100 N cm⁻¹. Spring free end is pulled to the right at 2 m s⁻¹; at t = 0 spring is unstretched and block moves left at 1 m s⁻¹. Find maximum spring extension (cm). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — In frame moving at 2 m s⁻¹ with the pulled end, initial block velocity is −1 − 2 = −3 m s⁻¹. Step 2 — Spring is initially natural length, so relative mechanical energy is (1/2)m(3)². Step 3 — At maximum extension relative velocity vanishes: (1/2)k xₘₐₓ² = (1/2)m × 9. Step 4 — k = 100 N cm⁻¹ = 10⁴ N m⁻¹ and m = 4 kg. Thus xₘₐₓ = 3√(4/10⁴) m = 0.06 m = 6 cm.