Question

A spring having a spring constant of 125 N/m hangs from a support and a 4.0 kg mass is attached to the other end. The mass is raised so that the spring is un-stretched and released from rest. How fast (in m/s) is the mass traveling after it has fallen 0.5 m?

Answer 1

Let x is the distance that the mass will be streched = 0.5 m

H is the initial height of the mass from groud

h is the final height of maa from ground

m be the mass = 4 kg

k be the spring constant = 125 N/m

let v be the velocity of the mass travelling after it has fallen to h

So x=H-h

Then by energy conservation :

$mgH+\frac{kx^{2}}{2}=mgh+\frac{mv^{2}}{2}$

$mg(H-h)+\frac{125 \times (0.5)^{2}}{2}=\frac{mv^{2}}{2}$

$\frac{4v^{2}}{2}=(4 \times 9.8 \times 0.5)+15.625$

$\frac{4v^{2}}{2}=19.6+15.625$

$2v^{2}=35.225$

$v^{2}=17.6125$

$\boldsymbol{v=4.2\;\;m/s\;(Answer)}$