Question

When a person of mass \( m \) climbs up or down a rope with uniform speed \( v \), the tension in the rope is ( \( g \) = acceleration due to gravity)

• A. \( mg \)
• B. \( m(g+v) \)
• C. \( m(g-v) \)
• D. \( mgv \)
• E. \( m\left(\frac{g}{v}\right) \)

Hint:

If velocity is constant, net force is zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem applies Newton's First Law of Motion. The key phrase is "uniform speed." Uniform speed means the velocity is constant, which in turn means the acceleration is zero.
Step 2: Key Formula or Approach:
According to Newton's First Law (or the special case of the Second Law where a=0), if the net force on an object is zero, its velocity is constant. \[ \sum \vec{F}_{net} = m\vec{a} = 0 \] We need to draw a free-body diagram for the person and apply this condition. The forces acting on the person are: 1. Gravity (\(F_g = mg\)), acting downwards. 2. Tension from the rope (\(T\)), acting upwards.
Step 3: Detailed Explanation:
The person is moving with a uniform speed `v`. This means the acceleration `a` is zero. The net force acting on the person must be zero. Let's consider the vertical forces. We can set the upward direction as positive. The upward force is the tension `T`. The downward force is the weight `mg`. \[ F_{net} = T - mg \] According to Newton's first law, since the acceleration is zero: \[ F_{net} = ma = m(0) = 0 \] Therefore: \[ T - mg = 0 \] \[ T = mg \] This result holds true whether the person is climbing up or down, as long as the speed is uniform (acceleration is zero).
Step 4: Final Answer:
The tension in the rope is mg.