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

Answer:

v = 1.30 m/s

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

 $\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2$

 $v = \sqrt{\dfrac{kx'^2}{m}}$

 $v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}$

 $v = \sqrt{1.6958}$

    v = 1.30 m/s