The concepts in this problem are similar to those in Multiple-Concept Example 4, except that the force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a 75.4-kg person straight up by means of a cable. The person has an upward acceleration of 0.683 m/s2 and is lifted from rest through a distance of 8.39 m. (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person's weight? (d) Use the work-energy theorem and find the final speed of the person.

To solve this problem, we can use the work-energy theorem. The work done on an object is equal to the change in its kinetic energy.

(a) To find the tension in the cable, we need to calculate the force exerted by the cable. We can use Newton's second law, F = ma, where F is the force, m is the mass of the person, and a is the person's upward acceleration.

F = ma
F = (75.4 kg)(0.683 m/s^2)
F = 51.4612 N

So, the tension in the cable is 51.4612 N.

(b) To find the work done by the tension in the cable, we can use the work formula, W = Fd, where W is the work done, F is the force, and d is the distance over which the force is applied.

W = Fd
W = (51.4612 N)(8.39 m)
W ≈ 431.6377 J

Therefore, the tension in the cable does approximately 431.6377 J of work.

(c) To find the work done by the person's weight, we need to calculate the gravitational potential energy. The formula for gravitational potential energy is PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

PE = mgh
PE = (75.4 kg)(9.8 m/s^2)(8.39 m)
PE ≈ 5960.6976 J

Therefore, the person's weight does approximately 5960.6976 J of work.

(d) According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. The initial kinetic energy is zero because the person starts from rest. Therefore, the work done by the tension in the cable and the work done by the person's weight equals the final kinetic energy.

Work done by the tension in the cable + Work done by the person's weight = Final kinetic energy

431.6377 J + 5960.6976 J = (1/2)mv^2

Let's solve for v:

v^2 = (431.6377 J + 5960.6976 J) x 2/m
v^2 ≈ (6392.3353 J) x 2/(75.4 kg)
v^2 ≈ 169.2990 m^2/s^2
v ≈ √(169.2990 m^2/s^2)
v ≈ 13.01 m/s

Therefore, the final speed of the person is approximately 13.01 m/s.