Question

A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an additional 0.15 m and then released. Determine the maximum speed of the mass.

Answer 1

v = 1.30 m/s

Step-by-step explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged (x)= 0.13 m

now,

weight of the mass attached = Kx

m g = k x

0.35 x 9.8 = k x 0.13

k = 26.38 N/m

now, using conservation of energy

`(1)/(2)mv^2 = (1)/(2)kx'^2`

`v = \sqrt{(kx'^2)/(m)}`

`v = \sqrt{(26.38 * 0.15^2)/(0.35)}`

`v = √(1.6958)`

v = 1.30 m/s