Question

A hanging rope can withstand a tension of 750 N. The minimum acceleration for a man of 50 kg to climb up without breaking the rope is $(g=10\text{ ms}^{-2})$

• A. $6\text{ ms}^{-2}$
• B. $4\text{ ms}^{-2}$
• C. $5\text{ ms}^{-2}$
• D. $7\text{ ms}^{-2}$
• E. $2\text{ ms}^{-2}$

Hint:

Logic Tip: The question phrasing says "minimum acceleration... without breaking," but physically, $5\text{ ms}^{-2}$ is the \textit{maximum} acceleration the man can have before the rope snaps. Any acceleration greater than 5 will exceed 750 N.

Answer 1

The Correct Option is C

Concept:
When a person climbs up a rope with an acceleration $a$, the rope must support both the person's weight and the additional force required to accelerate their mass upwards. The equation for the tension $T$ in the rope is given by Newton's Second Law: $$T - mg = ma \implies T = m(g + a)$$
Step 1: Identify the given parameters.
Maximum breaking tension, $T_{max} = 750\text{ N}$ Mass of the man, $m = 50\text{ kg}$ Acceleration due to gravity, $g = 10\text{ ms}^{-2}$
Step 2: Set up the tension inequality.
To ensure the rope does not break, the tension must be less than or equal to the maximum withstandable tension: $$T \le T_{max}$$ $$m(g + a) \le 750$$
Step 3: Solve for the acceleration (a).
Substitute the known values into the inequality: $$50(10 + a) \le 750$$ Divide both sides by 50: $$10 + a \le 15$$ Subtract 10 from both sides: $$a \le 5\text{ ms}^{-2}$$ Thus, the critical threshold for acceleration is $5\text{ ms}^{-2}$.