need help with physics please?

Question

A rescue helicopter lifts a 74 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s^2 and is lifted from rest through a distance of 13 m.

(a) What is the tension in the cable?
(b) How much work is done by the tension in the cable?
(c) How much work is done by the person's weight?
(d) Use the work-energy theorem and find the final speed of the person.

Answer 1

Ok, let's go on : Interesting problem

First let's use the Newton¡s second law :

Tension - Weight = mass*acceleration.

T - 74*9.8 = 74*0.7

T = 777 Newtons

b) Workd done by tension : Tension*distance = Work

W = 777*13 = 10101 Joules

c) Workd done by weight : Weight*distance = Work

W' = 74*9.8*13 = 9427.6 Joules

d) Theorem of work and energy :

Initial energy - final energy = Work done by friction

initial energy = 0 Joules >> initial kinetic energy

final energy = potential energy plus final kinetic energy

74*9.8*13 + 74*v^2 / 2 = (777*13)

v = 4.26 m/s

Note, the last part is tricky, because, you need to use the work by the tension, how did I figure it out ?

use this :

Vf^2 = 0^2 + 2*0.7*13

vf = 4.26 m/s

Hope that helps