NLM O3

Solution

We analyze the relationship between the slope of the Tension-Distance graph and the linear mass density of the rope.

Consider a rope of length $L$ being pulled by a force $F$ at $x=L$. The rope is on a frictionless floor. Let the acceleration of the rope be $a$. Consider a section of the rope from the rear end ($x=0$) to a distance $x$. Let the mass of this section be $m(x)$.

The tension $T(x)$ at distance $x$ is the force responsible for accelerating the mass behind it ($m(x)$).

$$ T(x) = m(x) \cdot a $$

Differentiating the tension with respect to $x$:

$$ \frac{dT}{dx} = \frac{dm(x)}{dx} \cdot a $$

The term $\frac{dm(x)}{dx}$ represents the linear mass density $\lambda(x)$ at that point.

$$ \frac{dT}{dx} = \lambda(x) \cdot a $$

Since the acceleration $a$ is constant for the whole rope, the slope of the $T$ vs $x$ graph is directly proportional to the linear mass density $\lambda(x)$.

Looking at the provided graph:

• The slope is initially steep (high value) near $x=0$.
• The slope decreases as $x$ increases (the curve becomes flatter).

Since $\text{Slope} \propto \text{Density}$, a decreasing slope implies a decreasing density.

Therefore, the density of the rope decreases with distance $x$ from the rear end.