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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.

Respuesta :

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

 [tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2[/tex]

 [tex]v = \sqrt{\dfrac{kx'^2}{m}}[/tex]

 [tex]v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}[/tex]

 [tex]v = \sqrt{1.6958}[/tex]

    v = 1.30 m/s

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