Question

A block of mass 5 kg is released from rest when compression in the spring is 2 m. Block is not attached with the spring and natural length of spring is 4m. Maximum height of block from ground is?

Answer 1

block is released from rest when compression in spring is 2m. so, potential energy of spring is U = 1/2 K(2m)²

here, K = 300 N/m

so, U = 1/2 × 300 × 4 = 600 J , this energy given to block. now block moves upward and gain sum speed.

so, potential energy of spring = potential energy due to height + kinetic energy

or, 600J = mgh + 1/2 mv²

here, h = 4m - 2m = 2m

so, 600 = 5 × 10 × 2 + 1/2 × 5 v²

or, 600 - 100 = 5/2 × v²

or, v² = 200 ⇒v = √200

now, vertical component of velocity = vsin30°.

at maximum height, block becomes rest.

so, 0² = (vsin30°)² + 2(-g)x

or, x = v²sin²30°/2g = 200 × (1/4)/2(10)

= 2.5m

hence, height of block = 2.5 + 6sin30°

= 2.5 + 3 = 5.5 m