Question

A 70 kg person in sea is being lifted by a helicopter with the help of a rope which can bear a maximum tension of 100 kg-weight

Answer 1

here apparent weight W= m(g+a)
100g = 70(g+a)
10g = 7g +7a
3g = 7a

a = 3 × 9.8
______ = 4.2 m/sec sq.
7

Answer 2

Hii there,
Question given here is incomplete.

● Complete question should be like this-
A 70 kg man in sea is being lifted by a helicopter with the help of a rope which can bear a maximum tension of 100kg-wt. With what maximum acceleartion the helicopter should rise so that the rope may not break.

● Answer-
a = 4.286 m/s^2

● Explanation-
# Given-
T = 100 kg-wt = 1000 N
m = 70 kg

# Solution-
Let a be the required acceleration.
Then, for maximum tension,
ma = T - mg
70a = 1000 - 70×10
70a = 300
a = 4.286 m/s^2

Therefore, maximum acceleration without breaking the rope is 4.286 m/s^2.

Hope this helps...