Asked by Ahmed mostafa
a mass of 2.1 kg attached to a verticalspring stretches the spring 6.2 cm from its original equilibriumposition, what is the spring constant?
Answers
Answered by
Anonymous
Weight in Newtons = m g = 2.1 * 9.81 = F
stretch in meters = 0.062 = x
F = k x
2.1 * 9.81 = k * 0.062
stretch in meters = 0.062 = x
F = k x
2.1 * 9.81 = k * 0.062
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