Question

A rescue helicopter lifts a 81.9-kg person straight up by means of a cable. The person has an upward acceleration of 0.646 m/s2 and is lifted from rest through a distance of 8.59 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. Please include all decimal points in answer! Thank you!

Answer 1

Newton's famous second law tells us that F=ma (Force equals mass times acceleration). In the case of the rescued person, m = 81.9 kg, and a = 0.646 m/s^2. Therfore, we have F = 52.9074N.

a) Only two forces act on the person here: gravity and tension in the cable. We know that the upward force F is equal to T-Fg. The person's weight is given by mg = 9.8 m/s^2 * 81.9 kg = 802.62 N. Tension in the cable must then be 855.5274 N so that F = 52.9074 N = 855.5274 N - 802.62N.

b) Work is given by the dot product of a force and a displacement. Since here both the force and displacement are in the same direction, we can simply multiply the two. For the person as well as the cable, the displacement is 8.59 m upward.

For the cable, we gave W = F * d = 855.5274 N * 8.59 m = 7348.980 J.

c) For the person, we'll put a negative sign in front of the force, since it's pointing in the opposite direction to the dispacement. W = F * d = -802.62 N * 8.59 m = - 6894.5058 J.

d) As far as energy is concerner, the person gained 7348.980 J from the tension in the cable, but 6894.5058 J were stored as potential gravitational energy, for a net kinetic energy gain of 454.4742 J. The variation in kinetic energy can be rewritten as ΔE = 1/2 m (Δv^2).

We can isolate v to get Δv = √(2ΔE/m) = √(2*454.4742J/81.9kg) ≈ 3.33 m/s. Assuming the initial speed was zero, the final speed will be 3.33 m/s.