Question

# When a 0.427 kg mass is hung from a certain spring it stretches 0.122 m to...

When a 0.427 kg mass is hung from a certain spring it stretches 0.122 m to its equilibrium position at point P. If this mass is pulled down 0.151 m from point P and released, what is the magnitude of the velocity in m/s of this mass 0.0392 m from point P? A 1.42 m3 piece of wood with a density of 0.805 kg/m3 floats in the ocean where the density of the water is 1.027 kg/m3. What is the volume of the wood that is submerged in m3 ?

Question : When a 0.427 kg mass is hung from a certain spring it stretches 0.122 m to its equilibrium position at point P. If this mass is pulled down 0.151 m from point P and released, what is the magnitude of the velocity in m/s of this mass 0.0392 m from point P?

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Solution :

Given :

m = 0.427 kg

x = 0.122 m

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Since: For equilibrium :

k x = mg

Thus : k = mg / x = (0.427)(9.81) / (0.122) = 34.335 N/m

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Now, here : A = 0.151 m

and, x = 0.0392 m

Thus, According to the conservation of energy : (1/2) m v2 + (1/2) k x2 = (1/2) k A2 v2 = (k / m) (A2 - x2) v2 = { (34.335) / (0.427) } {(0.151)2 - (0.0392)2} = 1.71 m2/s2 v = 1.308 m/s