A rescue helicopter is lifting a man (weight = 925 N) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 2.59 m/s2? (b) What is the tension during the remainder of the rescue when he is pulled upward at a constant velocity?

1) ma=T-mg

T=m(a+g)
2) a=0
=>
0=T-mg
T=mg

To find the tension in the cable when the man is given an initial upward acceleration and when he's pulled upward at a constant velocity, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).

(a) When the man is given an initial upward acceleration of 2.59 m/s^2:
We need to find the tension in the cable, which is the net force acting on the man.

First, we calculate the man's mass (m) using the formula weight = mass × acceleration due to gravity (W = mg). Weight is given as 925 N, and the acceleration due to gravity (g) is 9.81 m/s^2. Rearranging the formula, we have:

m = W / g = 925 N / 9.81 m/s^2 ≈ 94.18 kg

Now we can find the tension (T) using Newton's second law:

T = m × acceleration = 94.18 kg × 2.59 m/s^2 ≈ 244 N

Therefore, the tension in the cable when the man is given an initial upward acceleration of 2.59 m/s^2 is approximately 244 N.

(b) When the man is pulled upward at a constant velocity:
Since the man is pulled upward at a constant velocity, the acceleration is zero. According to Newton's second law, if the acceleration is zero, the net force acting on the man is also zero. Therefore, the tension in the cable is equal to the man's weight.

T = W = 925 N

Therefore, the tension in the cable during the remainder of the rescue when the man is pulled upward at a constant velocity is 925 N.