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Question
ive asked this before but no luck here goes, Suppose that 2J of work is
needed to
stretch a spring from its natural length of 30cm to a length of 42cm
a) how much work is needed to stretch the spring from 35cm to 40cm?
b) how far beyond its natural length will a force of 30N keep the spring
stretched?
answer: a)1.04J b)10.8cm
what ive tried:
since f(x) = kx i integrated from
.12{
| kx dx
0}
where k is the spring constant and x is the distance stretched past the natural
length. ive tried to use this to find k but i get it wrong please help.
The standard equation for a spring is F=-kx where
x is the distance that the spring has been stretched or compressed away from the equilibrium position, which is the position where the spring would naturally come to rest,
F is the restoring force exerted by the material [usually in newtons], and k is the force constant (or spring constant). The constant has units of force per unit length (usually in Newtons per meter). This comes from http://en.wikipedia.org/wiki/Hooke's_law
Using this, we have 0.5kx^2 (the integral of kx) evaluated from 0 to 0.12 gives 2 Joules. This says that 0.0072k=2, or k=277.89. Evaluating the intgral from 0.05 to 0.10 gives 138.89(0.0100-0.0025)=1.04.
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