Question

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 77.0-kg person straight up by means of a cable. The person has an upward acceleration of 0.730 m/s² and is lifted from rest through a distance of 14.1 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.

I got:

A.)810.81 N

B.)11432.42 J

C.)-10639.86 J

D.)45.37 m/s

Not sure If the last one is right. Can anyone give me an insight?

Answer 1

A)

T = tension force acting in upward direction

m = mass of the person = 77 kg

W = weight of the person in down direction = mg = 77 x 9.8 = 754.6 N

a = acceleration in upward direction = 0.730 m/s²

Force equation for the motion of the person is given as

T - W = ma

T - 754.6 = 77 (0.730)

T = 810.81 N

B)

d = distance through which person is lifted = 14.1 m

Work done by tension force is given as

W_t = T d Cos0

W_t = (810.81) (14.1) Cos0

W_t = 11432.42 J

C)

Work done by persons weight is given as

W_g = W d Cos180

W_g = (754.6) (14.1) Cos180

W_g = -10639.86 J

D)

F_net = net force acting on the person = ma

v_o = initial velocity = 0 m/s

v_f = final velocity = ?

Using work-change in kinetic energy theorem

F_net d = (0.5) m (v_f² - v_o²)

ma d = (0.5) m (v_f² - v_o²)

(0.730) (14.1) = (0.5) (v_f² - 0²)

v_f = 4.5 m/s