Question

A spring (k = 100 Nm^-1) is suspended in vertical position having one end fixed a
with a 2kg block. When the spring is in non deformed shape, the block is given initial velocity 2m/s in
downward direction. The maximum elongation of the spring is
​

Answer 1

Dear Student,

◆ Answer -

Maximum elongation = 28.28 cm

● Explaination -

# Given -

k = 100 N/m

m = 2 kg

v = 2 m/s

# Solution -

When spring is deformed, kinetic enrgy of mass is converted to potential energy of spring.

mv²/2 = kx²/2

x² = mv²/k

x = v √(m/k)

Now, substitute values,

x = 2 √(2/100)

x = 2√2 / 10

x = 0.2828 m

x = 28.28 cm

Therefore, the maximum elongation of the spring is 28.28 cm.

Thanks dear. Hope this helps you...