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Chapter 5: Problem 40

A spring with spring constant \(k\) is initially compressed a distance \(x_{0}\) from its equilibrium length. After returning to its equilibrium position, the spring is then stretched a distance \(x_{0}\) from that position. What is the ratio of the work that needs to be done on the spring in the stretching to the work done in the compressing?

Short Answer

Expert verified
Answer: The ratio of the work done during stretching to the work done during compression is \(-1\).

Step by step solution

01

Calculate work done during compression

During compression, the spring is initially compressed a distance \(x_{0}\). The work done, \(W_{c}\), can be calculated as the integral of the force over the distance compressed: $$ W_{c} = \int_0^{x_0} F dx_c$$ For a spring, the force \(F\) is given by the Hooke's law mentioned above: $$ F = -kx_c$$ Now we plug \(F\) into the integral and solve for \(W_{c}\): $$ W_{c} = \int_0^{x_0} -kx_c dx_c$$ $$ W_{c} = -\frac{1}{2} kx_0^2$$
02

Calculate work done during stretching

Now we calculate the work done during stretching, \(W_{s}\). The spring is initially at its equilibrium position, and it is stretched a distance \(x_0\). The work done in this case is: $$ W_{s} = \int_0^{x_0} F dx_s$$ When the spring is stretched, the force is given by Hooke's law: $$ F = kx_s$$ Now we plug \(F\) into the integral and solve for \(W_{s}\): $$ W_{s} = \int_0^{x_0} kx_s dx_s$$ $$ W_{s} = \frac{1}{2} kx_0^2$$
03

Find the ratio of work done

The ratio of the work done during stretching to the work done during compression is: $$ R = \frac{W_{s}}{W_{c}}$$ Substituting the values we found for \(W_{s}\) and \(W_{c}\) into the equation, we get: $$ R = \frac{ \frac{1}{2} kx_0^2}{-\frac{1}{2} kx_0^2}$$ $$ R = -1$$ Therefore, the ratio of the work done during stretching to the work done during compression is \(-1\).

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