A block of mass 2Kg is attached to the lower end of a spring of force constant 1960 Nm−1 hanging from rigid support. Initially spring is unstretched, if the block is suddenly released the maximum elongation produced in spring is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

1cm

2cm

2.2cm

3cm

answer is B.

(Unlock A.I Detailed Solution for FREE)

Detailed Solution

Loos of P.E= gain of elastic potential energymg⁡x=12kx2x=2mgk=2×2×9.81960=2100mx=2cm

Watch 3-min video & get full concept clarity