Question

A man weighing 100 kg slides down a light rope with an acceleration of 1.8 m s\(^{-2}\). If \(g = 9.8\) m s\(^{-2}\), the tension in the rope is

• A. 180 N
• B. 1160 N
• C. 800 N
• D. weightlessness

Hint:

For sliding down: \(T = m(g-a)\). For climbing up with acceleration: \(T = m(g+a)\). Always define a clear positive direction.

Answer 1

The Correct Option is C

Step 1: Basic Principle
When sliding down with acceleration \(a\), Newton's second law: \(mg - T = ma\).
Step 2: Solution Procedure:
\[ T = m(g - a) = 100 \times (9.8 - 1.8) = 100 \times 8 = 800 \text{ N} \]
Step 3: Required Answer:
Tension in the rope is 800 N.