A massless spring with a 1.5 kg mass attached is suspended vertically, an additi

Question

A massless spring with a 1.5 kg mass attached is suspended vertically, an additional force of -10N is applied to the system. The overall displacement of the spring from its unstrained equilibrium position is 50 cm. a) Determine the spring constant. b) Determine the equilbrium position of the spring when only the weight is attached. In other words the position of the spring when the weight is just haninging on the spring and not moving. c) Determine the velocity of the 1.5 kg mass when it passes through the equilibrium position of the spring with the weight attached, when it is traveling upward.

Explanation / Answer

part A: Spring constant K is obtanined from

F= Kx

K = 10/0.5

K = 20 N/m
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use F = mg = kx

x = 1.5* 9.8/20

x = 0.735 m

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use Energy KInetic = EPE

0.5mv^2 = 0.5 kx^2

v^2 20 * 0.735 * 0.735/1.5

V = 2.68 m/s

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