Rope Tension Physics Problem Explained

This problem involves calculating the tension in a rope being pulled horizontally across a smooth floor. We need to find the tension at a specific point, 6 m from where the force is applied.

Calculating Acceleration

First, find the total mass of the rope and then its acceleration.

• Rope Length ($L$) = 10 m
• Linear Density ($μ$) = 0.5 kg/m
• Applied Force ($F$) = 25 N
• Total Mass ($M$) = $L \times μ = 10 \text{ m} \times 0.5 \text{ kg/m} = 5 \text{ kg}$
• Using Newton's second law, $F = Ma$, the acceleration ($a$) is: $a = \frac{F}{M} = \frac{25 \text{ N}}{5 \text{ kg}} = 5 \text{ m/s}^2$

Determining Tension

The tension at a point in the rope is the force required to accelerate the segment of the rope beyond that point (the part being pulled). We need the tension 6 m from the point of application.

• The length of the rope segment being accelerated at the 6 m mark is the remaining length: $10 \text{ m} - 6 \text{ m} = 4 \text{ m}$.
• Calculate the mass ($m$) of this 4 m segment: $m = \text{Length} \times μ = 4 \text{ m} \times 0.5 \text{ kg/m} = 2 \text{ kg}$
• The tension ($T$) at the 6 m point is the force needed to accelerate this 2 kg mass: $T = m \times a = 2 \text{ kg} \times 5 \text{ m/s}^2 = 10 \text{ N}$

Therefore, the tension in the rope at a point 6 m away from the point of application is 10 N.