It takes 80J of work to stretch a spring 0.5m from its equilibrium position. How much work is needed to stretch it an additional .5m?

Question

It takes $80\,\textrm{J}$ of work to stretch a spring $0.5\,\textrm{m}$ from its equilibrium position. How much work is needed to stretch it an additional $0.5\,\textrm{m}$?

Here's what I have: $F(x)=kx$ and $F(0.5) = k(80)$. Solving for $k$, I get $k = 160$. Then, I find that $\int_{0.5}^1 160x dx = 60$, but that's not an answer..

Answer 1

You are confusing force and work in the first step. Work is measured in joules, so your given quantity is work. Since you are stretching 0.5 meters from equilibrium position, and work is the integral of the force used over the distance moved, we have that

$$ \mathbf{80} = \int_0^{0.5} F(x) \, dx = \int_0^{0.5} kx \, dx.$$ Then solve this for $k$, then integrate from $0.5$ to $1$ the given quantity.

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More details, in response to questions: Once we have the correct* number in the left-hand side, we have that

\begin{align} 80 &= \int_0^{0.5} kx \, dx\\ & = k \int_0^{0.5} x \, dx\\ & = k \left[ \frac{x^2}{2} \right]_{x = 0}^{x = 0.5}\\ & = k \left( \frac{(0.5)^2}{2} - \frac{0^2}{2} \right) \\ & = k \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 =\frac{k}{8}. \end{align}

Thus, $k = 80 \cdot 8 = 640$. Then continue as you did.

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In general, for these problems, you are given a certain (force/work) and asked to find another (force/work). That's 4 possibilities, and you should know how to handle any combination.

Answer 2

Integrating $F(x)=kx$ we get the formula for the elastic energy stored in a spring (=the work to stretch the spring from equilibrium):

$$W=\frac 12kx^2$$

Use this to find the spring constant $k$, then the energy needed to stretch to the new position, then subtract the two energies to find the needed work.

This is how I teach this kind of problem in my $12$th grade physics class.

Answer 3

The potential energy of a string when stretched $x$ meters from its equilibrium position is:

$$ U(x) = -\frac{1}{2}kx^2 $$

The work done when stretching a spring from points $x_1$ to $x_2$ is just the difference of potential energy

$$ W_{x_1 \rightarrow x_2} = U(x_2) - U(x_1) $$

From this you have the condition:

$$U(0.5) = 80$$

And you need to find

$$ U(1) - U(0.5) $$