1. Understand the setup

A man of mass $m=60\,\text{kg}$ climbs a light rope passing over a frictionless, massless pulley.

• One end is fixed to hook $C$.
• The other end is the free end on which the man climbs.
• Since the rope and pulley are ideal, the tension is the same throughout the rope.
• Maximum tension the rope can bear is $T_{\max}=360\,\text{N}.$

We need the maximum safe upward acceleration of the man.

2. Forces on the man

When the man climbs the rope, the rope exerts an upward tension $T$ on him. His weight acts downward: $mg=60g.$

If he accelerates upward with acceleration $a$, then applying Newton's second law upward: $T-mg=ma.$

So, $T=m(g+a).$

3. Use the safety condition

For maximum safe acceleration, tension must not exceed $360\,\text{N}$: $m(g+a)\le 360.$

Substitute $m=60\,\text{kg}$ and take $g=10\,\text{m/s}^2$: $60(10+a)\le 360.$

Divide by $60$: $10+a\le 6.$

This gives $a\le -4\,\text{m/s}^2.$

This is impossible for upward climbing, so clearly the above interpretation misses the pulley effect.

4. Correct interpretation of the system

Since one end of the rope is fixed and the man climbs on the other side over the pulley, the man is effectively supported by two upward tensions through the rope arrangement.

Thus the upward force on the man is $2T.$

Applying Newton's second law upward: $2T-mg=ma.$

So, $2T=m(g+a).$

Hence, $T=\frac{m(g+a)}{2}.$

5. Apply maximum tension condition

$\frac{m(g+a)}{2}\le 360$

Substitute $m=60$ kg: $\frac{60(10+a)}{2}\le 360$ $30(10+a)\le 360$ $10+a\le 12$ $a\le 2\,\text{m/s}^2.$

This still does not match the options, so let us examine the standard climbing-rope relation more carefully.

6. Standard result for a man climbing a rope over a pulley with the other end fixed

If the man climbs with acceleration $a$ relative to ground, the tension in the rope is determined from the man's free body diagram alone: $T-mg=ma.$

Thus, $T=m(g+a).$

Using $T_{\max}=360$ N gives $60(10+a)=360$ $10+a=6$ $a=-4\,\text{m/s}^2.$

This again is inconsistent with upward climbing, meaning the intended answer in such problems usually assumes the man is pulling the rope so that the rope tension acts twice in the full support arrangement, and with $g=9.8\approx 10$ one still does not get the option.

7. Check option-based consistency

For option $a=4$: $T=\frac{60(10+4)}{2}=420\,\text{N}$ which exceeds $360$ N.

So physically, option $4$ is not correct under the ideal pulley-rope analysis.

For safe tension $360$ N, the acceleration from the two-tension model is only $a=2\,\text{m/s}^2.$

Since $2$ is not among the options, the closest intended textbook answer may have been obtained by using a different convention or there is an issue in the problem/options.

8. Comparison with stored answer

My derived result is that the maximum safe acceleration should be $2\,\text{m/s}^2,$ not $4\,\text{m/s}^2$.

Therefore I disagree with the stored correct answer.