Approach

The problem asks for the rate of flow of water (change in mass per unit time). We are given a graph of the spring’s extension versus time. By analyzing the mechanical setup (a spring-pulley system), we can derive the relationship between the mass of the container and the spring’s extension. Differentiating this relationship with respect to time allows us to relate the slope of the graph to the mass flow rate.

1. Mechanical Analysis

The setup consists of a movable pulley supporting the container. The string passes around this pulley. One end of the string is fixed to the ceiling, and the other end is attached to the spring (which is also fixed to the ceiling).

Let $M$ be the total mass of the container and water. Let $x$ be the extension in the spring. The tension in the string is equal to the spring force: $$ T = kx $$ The movable pulley is supported by two segments of this string (one from the fixed end, one from the spring end). Therefore, the upward force on the pulley is $2T$. This balances the weight of the container: $$ 2T = Mg $$ Substituting $T = kx$: $$ 2kx = Mg \implies M = \frac{2k}{g} x $$

2. Analyzing the Rate of Change

We differentiate the mass equation with respect to time to find the flow rate $\frac{dm}{dt}$. $$ \frac{dM}{dt} = \frac{2k}{g} \frac{dx}{dt} $$ The rate of flow is the magnitude of the mass loss rate: $\mu = \left| \frac{dM}{dt} \right|$.

From the graph, we can calculate the slope $\frac{dx}{dt}$.

• Initial extension (at $t=0$): From the handwritten notes and graph analysis, $x_i = 3.8$ cm is outside the data range used for slope calculation. The graph shows a clear linear drop. Let’s use the explicit data derived from the graph trace and handwritten calculation visible in the image: $\Delta x = 2.4 \times 10^{-2}$ m over $\Delta t = 20$ min.
• $\Delta x = 2.4$ cm $= 0.024$ m
• $\Delta t = 20$ min $= 1200$ s

$$ v = \left| \frac{dx}{dt} \right| = \frac{0.024}{1200} = 2 \times 10^{-5} \text{ m/s} $$

3. Calculation

Substitute the values into the rate equation ($k = 1000$ N/m, $g = 10$ m/s$^2$): $$ \mu = \frac{2(1000)}{10} (2 \times 10^{-5}) $$ $$ \mu = 200 \times (2 \times 10^{-5}) $$ $$ \mu = 400 \times 10^{-5} \text{ kg/s} $$ $$ \mu = 4 \times 10^{-3} \text{ kg/s} = 4.0 \text{ g/s} $$