Question:

A spring is stretched after 4.0-kg object is hung on it vertically. The displacement is measured as 0.020 m. Then, one stretches it farther by 0.040 m. Find the work done by this additional force.

Answer:

Work can be obtained by integrating the force \(|F|=kx\) with respect to the correspondent displacement.

\[  W = \int^x_0 F\cdot dx  \]

\[  = \int^x_0 kx dx   = \frac{1}{2}kx^2  \]

In order to calculate the work, we need to find the spring constant,\(k\). Consider the free-body diagram. The spring and gravitational forces act on the hanging mass. According to the equation of motion, we have the net force:

\[  \sum F = kx – mg = ma = 0  \]

This system is in equilibrium, so \(a=0\). Thus,

\[  kx = mg  \]

\[  \rightarrow k = \frac{mg}{x} = \frac{4.0\cdot 9.8}{0.02}=1960 \ \mathrm{N/m}  \]

Now, we can calculate the work done by the external force.

\[  W = \frac{1}{2}kx^2  \]

\[    = \frac{1}{2}1960 \cdot 0.04^2  \]

\[    = 1.6 \ \mathrm{J}  \]