When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.78 cm.
(a) What is the force constant of the spring?
N/m
(b) If the 2.00-kg object is removed, how far will the spring
stretch if a 1.00-kg block is hung on it?
cm
(c) How much work must an external agent do to stretch the same
spring 9.00 cm from its unstretched position?
Gravitational acceleration = g = 9.81 m/s2
Force constant of the spring = k
Mass of the first object = m1 = 2 kg
Extension of the spring due to the first object = X1 = 2.78 cm = 0.0278 m
m1g = kX1
(2)(9.81) = k(0.0278)
k = 705.755 N/m
Mass of the second object = m2 = 1 kg
Extension of the spring due to the second object = X2
m2g = kX2
(1)(9.81) = (705.755)X2
X2 = 0.0139 m
X2 = 1.39 cm
Work done to stretch the spring by 9 cm = W
X = 9 cm = 0.09 m
W = kX2/2
W = (705.755)(0.09)2/2
W = 2.858 J
a) Force constant of the spring = 705.755 N/m
b) Amount the spring will stretch due to the 1 kg block = 1.39 cm
c) Work done by an external agent to stretch the spring by 9 cm = 2.858 J
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