Work done on elastic springs, and Hooke's law
The work required to stretch or compress a spring
To calculate the work done when we stretch or compress an elastic spring, we’ll use the formula
???W=\int^b_aF(x)\ dx???
where ???W??? is the work done, ???F(x)??? is the force equation, and ???[a,b]??? is the distance over which the spring is stretched or compressed.
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Every spring has its own spring constant ???k???. This spring constant is part of Hooke’s Law, which states that
???F(x)=kx???
where ???F(x)??? is the force required to stretch or compress the spring, ???k??? is the spring constant, and ???x??? is the difference between the natural length and the stretched or compressed length. Since ???k??? is unique to each spring, we’ll need to calculate it prior to determining work, unless it’s given in the problem.
Keep in mind that we’ll want to find work in terms of Joules J, which is the same as Newton-meters N-m.
Video example of work done on elastic springs
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When you know natural length and the force required to stretch the spring
Example
A spring has a natural length of ???30??? cm. A ???50??? N force is required to stretch and hold the spring at a length of ???40??? cm.
How much work is done to stretch the spring from ???42??? cm to ???48??? cm?
How much work is done to compress the spring from ???30??? cm to ???25??? cm?
We’ll use Hooke’s Law to find ???F(x)???, but first we need to find ???k???.
Since we know that a ???50??? N force is required to stretch and hold the spring at a length of ???40??? cm, from its natural length of ???30??? cm, we’ll set ???F(x)=50??? and ???x=0.10??? m, which is the difference between the natural length and the stretched length, converted from cm to m. Remember that we’ll be finding work in terms of Newtons and meters, which is why we converted ???10??? cm to ???0.10??? m.
???50=0.10k???
???k=500???
With ???k???, we can develop a generic equation for our spring using Hooke’s Law.
???F(x)=kx???
???F(x)=500x???
Every spring has its own spring constant K. This spring constant is used in the Hooke’s Law formula.
Work done to stretch the spring
To calculate the work required to stretch the spring from ???42??? cm to ???48??? cm, we pretend that the spring at its natural length of ???30??? cm ends at the origin, which means that stretching it to ???42??? cm means we’ve stretched it to ???12???, because ???42-30=12???. Stretching it to ???48??? cm means we’ve stretched it from the origin to ???18???, because ???48-30=18???.
Therefore, the work equation would be
???W=\int^b_aF(x)\ dx???
???W=\int^{18}_{12}500x\ dx???
But we need to convert the units from cm to m, so the interval becomes ???0.12??? m to ???0.18??? m.
???W=\int^{0.18}_{0.12}500x\ dx???
???W=250x^2\Big|^{0.18}_{0.12}???
???W=250(0.18)^2-250(0.12)^2???
???W=4.5???
The work done to stretch a spring with natural length ???30??? cm and spring constant ???k=500??? from ???42??? cm to ???48??? cm is ???4.5??? J.
Work done to compress the spring
To calculate the work required to compress the spring from ???30??? cm to ???25??? cm, we pretend that the spring ends at the origin, which means that compressing it to ???25??? cm means we’ve compressed it to ???-5???, because ???25-30=-5???.
Therefore, the work equation would be
???W=\int^b_aF(x)\ dx???
???W=\int^{-5}_0500x\ dx???
But we need to convert the units from cm to m, so the interval becomes ???0??? m to ???-0.05??? m.
???W=\int^{-0.05}_0500x\ dx???
???W=250x^2\Big|^{-0.05}_0???
???W=250(-0.05)^2-250(0)^2???
???W=0.625???
The work done to compress a spring with natural length ???30??? cm and spring constant ???k=500??? from ???30??? cm to ???25??? cm is ???0.625??? J.
