Question

A monkey of mass 50 kg climbs on a rope which can withstand the tension \( T = 350 \, \text{N} \). If the monkey initially climbs down with an acceleration of \( 4 \, \text{m/s}^2 \) and then climbs up with an acceleration of \( 5 \, \text{m/s}^2 \), choose the correct option (\( g = 10 \, \text{m/s}^2 \)):

• A. \( T = 700 \, \text{N} \) while climbing upward
• B. \( T = 350 \, \text{N} \) while going downward
• C. Rope will break while climbing upward
• D. Rope will break while going downward

Hint:

When calculating tension for vertical motion, use \( T = m(g + a) \) for upward motion and \( T = m(g - a) \) for downward motion. Compare the result with the breaking strength to determine if failure occurs.

Answer 1

The Correct Option is C

Step 1: Upward ascent. The rope tension during upward movement is determined by \( T = m(g + a) \). Given \( m = 50 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), and \( a = 5 \, \text{m/s}^2 \), the calculation is \( T = 50 (10 + 5) = 750 \, \text{N} \). This tension exceeds the rope's breaking strength of \( 350 \, \text{N} \), causing it to break during upward motion.

Step 2: Downward descent. The rope tension during downward movement is given by \( T = m(g - a) \), with \( a = 4 \, \text{m/s}^2 \). Substituting values: \( T = 50 (10 - 4) = 300 \, \text{N} \). This tension is below the breaking strength of \( 350 \, \text{N} \), so the rope will not break during downward motion.

Final Answer: \[ \boxed{\text{Rope will break while climbing upward.}} \]