Solution: Climber Sliding Down a Rope

1. Physics Principles

We apply the Impulse-Momentum Theorem. The change in the climber’s momentum is equal to the integral of the net force over time.

$$ \int F_{net} dt = m \Delta v $$

The forces acting on the climber are gravity ($mg$) downwards and the rope tension/friction ($T$) upwards. Thus, $F_{net} = mg – T$.

2. Visual Analysis

3. Calculation

Given: Mass $m = 65$ kg, $g = 10 \text{ m/s}^2$. Total weight $W = 650$ N.

The net impulse is:

$$ J_{net} = \int_{0}^{2} (mg – T) dt = \int_{0}^{2} mg \, dt – \int_{0}^{2} T \, dt $$

Geometrically, this integral represents the area between the line $y=650$ (gravity) and the curve $T(t)$.

This area forms a triangle pointing downwards from the $650$ N line:

• Base: $\Delta t = 2.0$ s.
• Height (depth of dip): $650 – 325 = 325$ N.

$$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} $$ $$ J_{net} = \frac{1}{2} \times 2.0 \times 325 = 325 \text{ N}\cdot\text{s} $$

Now, equate impulse to change in momentum:

$$ m \Delta v = 325 $$ $$ 65 \Delta v = 325 $$ $$ \Delta v = \frac{325}{65} = 5 \text{ m/s} $$