A spring which obeys Hooke's law stretches 10 inches when 100 pounds is suspended from it. How far will it stretch when 120 pounds is suspended from it?
Hooke's Law states that the stretch of a spring is directly proportional to the weight suspended.
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Hooke's Law for springs states that the amount a spring is stretched from its relaxed length is directly proportional to the weight it is supporting:
W = kX or
`W/X = k` where w is the weight supported and x is the amount stretched from the rest length. k is known as Hooke's constant and is related to the stiffness of the spring.
To determine the answer to this question, we can approach it in at least two ways. a) through proportions, or b) through solving for Hooke's constant.
Because k is a constant for any given spring, the ratio of the weigh to the length will always equal k even when you change the weight. Therefore
`(W_(1))/(X_(1)) = k` and `(W_2)/(X_2) = k` By the transitive law of equality
`W_1/X_1 = W_2/X_2 rArr X_2 = (W_2 X_1)/W_1` or `X_2 = ((120)10)/(100) = 12` inches
b) `k = W_1/X_1 = 100/10 = 10` pounds/inch and is the same for all weights applied to the spring, so
`W_2/X_2 = 10 rArr X_2 = W_2/10 = 120/10 = 12` inches
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