Calculus Spring problem

Question

I am having problem understanding a problem. It goes as follows: A spring needs $750$ lbs of force to compress it $3$ inches. How much work for another $3$ inches of compression?

My first attempt:

$$0.25k=750\\ k=3000\\ \int _{0.25}^{0.5}\:3000x\,\mathrm dx=1500[(0.5)^2-(0.25)^2]=281.25.$$

I have tried this using the inches, instead of decimal feet; I get different answers. One guy stated on a different thread (not mine) that I should start with $W=\frac{1}{2}kx^2∣^b_a $ . I followed this and got $562.5$, which is not enough.

This can't be right; the answer should be more than $750$, no?

Answer 1

Consider F= -kx ,where k is the spring constant. When x = 3 inches ,F = 750 lbs.

so $750$ $\,=$ $\mathtt-k.3$

$\mathbf k$ = $\mathtt{\frac {-750}3 }$

The work done W = $\int F \,dx$ = $\int -k.x\,dx$.

In this case W = $\int_3^6 -(\frac{-750}{3}).x\,dx$

Which evaluates out to 3375 inch pounds of work.

Edit: had the wrong unit for the final answer

Answer 2

You calculated the work correctly.

The force $\boldsymbol{F}$ acting on the spring is given by the equation $\boldsymbol{F} = k\boldsymbol{x}$, where $k$ is the spring constant and $\boldsymbol{x}$ is the displacement from equilibrium. The spring constant is found by dividing the magnitude of the force by the magnitude of the displacement. \begin{align*} k & = \frac{F}{x}\\ & = \frac{750~\text{lb}}{0.25~\text{ft}}\\ & = 3000~\frac{\text{lb}}{\text{ft}} \end{align*} Therefore, the work done on the spring when it is compressed from $3$ inches to $6$ inches is \begin{align*} W & = \int_{0.25~\text{ft}}^{0.5~\text{ft}} F(x)~dx\\ & = \int_{0.25~\text{ft}}^{0.5~\text{ft}} 3000~\frac{\text{lb}}{\text{ft}} x~dx\\ & = 3000~\frac{\text{lb}}{\text{ft}} \int_{0.25~\text{ft}}^{0.5~\text{ft}} x~dx\\ & = 3000~\frac{\text{lb}}{\text{ft}} \cdot \frac{1}{2}x^2 \bigg|_{0.25~\text{ft}}^{0.5~\text{ft}}\\ & = 1500~\frac{\text{lb}}{\text{ft}} x^2 \bigg|_{0.25~\text{ft}}^{0.5~\text{ft}}\\ & = 1500~\frac{\text{lb}}{\text{ft}} (0.25~\text{ft}^2 - 0.0625~\text{ft}^2)\\ & = 1500~\frac{\text{lb}}{\text{ft}} (0.1875~\text{ft}^2)\\ & = 281.25~\text{lb}~\text{ft} \end{align*}

This can't be right; the answer should be more than 750, no?

Remember that $750~\text{lb}$ is the force that is applied to the spring to compress it by $3~\text{in}$. The quantity $281.25~\text{lb}~\text{ft}$ is the work done on the spring to compress it by an additional $3$ inches when the same force is applied. Work is a measure of energy, not force, which is why the units, which you omitted, are different. Since the units of energy and force are different, we cannot meaningfully compare their magnitudes. What we can do is compare the work done to compress the spring $3$ inches with the work done to compress it $6$ inches or from $3$ inches to $6$ inches.