Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an additional 3 inches.
My attempts at this problem have led to a dead end. Here is one of the four that I have tried.
$$ 18=k(4) $$
$$k=\frac{18}{4}$$
Now I have,
$$ F(x)=\frac{18}{4}x$$
Because the spring is stretched from $x=4$ to $x=7$, the work done is
$$W=\int_{4}^{7}\frac{18}{4}x dx$$
$$=110.25-36$$
$$W=74.25$$
This converted between foot-pounds and inch-pounds still yields the wrong answer. The correct answer is $37.125 \mathrm {foot-pounds}$.
Please, if anyone knows a correct way to achieve the answer using calculus and Hooke's law please respond.
2 Answers
$\begingroup$The error occurred when you used $18 = k(4)$, an incorrect assumption. Let's derive an equation relating the work needed.
We start with the definition of work:
$W = \int_a^bF_{spring}dx$
By Hooke's Law, we have:
$F_{spring} = kx$
So, to find the work needed, we have to evaluate the integral:
$W = \int_a^bF_{spring}dx = \int_a^bkxdx = \frac{1}{2}kx^2\mid_a^b$
Since $a = 0$,
$W_4 = 18= \frac{1}{2}k(4)^2$ and it turns out that $k = \frac{9}{4}$
Your value for $k$ is 2x the correct value, which explains why your answer is 2x the correct answer.
$\endgroup$ 1 $\begingroup$Everything above is correct except in the beginning you want to convert your bounds from inches to feet as follows:
4 inches = 1/3 feet
7 inches = 7/12 feet
I hope this helps
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