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: Climbing upward The tension in the rope during upward motion is calculated using the formula: \[ T = m(g + a), \] where: \( m = 50 \, \text{kg} \) (mass of the monkey), \( g = 10 \, \text{m/s}^2 \) (gravitational acceleration), \( a = 5 \, \text{m/s}^2 \) (acceleration while climbing upward). Substituting the values: \[ T = 50 (10 + 5) = 750 \, \text{N}. \] Since the tension \( T \) exceeds the rope's breaking strength (\( 350 \, \text{N} \)), the rope will break during upward motion. Step 2: Climbing downward The tension in the rope during downward motion is given by: \[ T = m(g - a), \] where \( a = 4 \, \text{m/s}^2 \) (acceleration while climbing downward). Substitute the values: \[ T = 50 (10 - 4) = 300 \, \text{N}. \] Here, the tension \( T \) is less than the breaking strength (\( 350 \, \text{N} \)), so the rope will not break during downward motion. Final Answer: \[ \boxed{\text{Rope will break while climbing upward.}} \]