Question

A string tied on a roof can bear a maximum tension
of 50 kg wt. The minimum acceleration that can
be acquired by a man of 98 kg to descend will be
[Take g = 9.8 m/s21
(1) 9.8 m/s2
(2) 4.9 m/s2 O9
(3) 4.8 m/s2
(4) 5 m/s2​

Answer 1

Answer:4.8m/s^2

Explanation:

50 kg wt = 50 × 9.8 N

T max. = 50 × 9.8 N

Weight of the man = mg = 98 × 9.8 N

As he moves downwards :

mg - T = ma

(98 × 9.8) - (50 × 9.8) = 98a

9.8 ( 98 - 50) = 9.8 × 10a

48 / 10 = a

Therefore, a = 4.8m/s^2

HOPE IT HELPED....

Answer 2

The maximum tension the string can bear= 50 Kg. Wt = 50 × 9.8N

Mass of the weird human hanging there = 98 Kg.

Now the force applied by the human must be either equal to the tension or less than than the tension if it exceed that the string will break hurting thr man which we don't want to happen.

So,

Let's make an equation fot it assuming the acceleration of the human to be "a",

Now,
- mg +T = - ma,
(consider the FBD in the attachment.)

Multiplying both sides with -ve 1.
mg - T = ma




Substiting thae values we have,

(98)(9.8) N - (50)(9.8)N = 98a N

960.4 - 490 = 98a

98a = 470.4

a = 4.8m/s^2

Hence option C