WORK DONE ON A SPRING SCALE
A woman weighing 600 N steps on a bathroom scale that contains a stiff spring (Fig. 6.21). In equilibrium, the spring is compressed 1.0 cm under her weight. Find the force constant of the spring and the total work done on it during the compression.
IDENTIFY and SET UP:
In equilibrium the upward force exerted by the spring balances the downward force of the woman’s weight. We’ll use this principle and Eq. (6.8) (F_{x}=k x) to determine the force constant k, and we’ll use Eq. (6.10) (W=\int_{x_{1}}^{x_{2}} F_{x} d x=\int_{x_{1}}^{x_{2}} k x d x=\frac{1}{2} k x_{2}^{2}-\frac{1}{2} k x_{1}^{2}) to calculate the work W that the woman does on the spring to compress it. We take positive values of x to correspond to elongation (upward in Fig. 6.21), so that both the displacement of the end of the spring (x) and the x-component of the force that the woman exerts on it (F_x) are negative. The applied force and the displacement are in the same direction, so the work done on the spring will be positive.
EXECUTE:
The top of the spring is displaced by x = -1.0 cm = -0.010 m, and the woman exerts a force F_x = -600 N on the spring. From Eq. (6.8) (F_{x}=k x) the force constant is then
k=\frac{F_{x}}{x}=\frac{-600 \mathrm{~N}}{-0.010 \mathrm{~m}}=6.0 \times 10^{4} \mathrm{~N} / \mathrm{m}Then, using x_1 = 0 and x_2 = -0.010 m in Eq. (6.10) (W=\int_{x_{1}}^{x_{2}} F_{x} d x=\int_{x_{1}}^{x_{2}} k x d x=\frac{1}{2} k x_{2}^{2}-\frac{1}{2} k x_{1}^{2}), we have
W=\frac{1}{2} k x_{2}^{2}-\frac{1}{2} k x_{1}^{2}=\frac{1}{2}\left(6.0 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)(-0.010 \mathrm{~m})^{2}-0=3.0 \mathrm{~J}
EVALUATE: The work done is positive, as expected. Our arbitrary choice of the positive direction has no effect on the answer for W.
You can test this by taking the positive x-direction to be downward, corresponding to compression. Do you get the same values for k and W as we found here?