Question

The person has an upward acceleration of 0.70 m/s^2 and is lifted from rest through a distance of 13 m. A rescue helicopter lifts a 80-kg person straight up by means the cable? (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

a)

840 N

b)

10920 J

c)

- 10192 J

d)

4.3 m/s

Step-by-step explanation:

a)

T = tension force in the cable in upward direction = ?

a = acceleration of the person in upward direction = 0.70 m/s²

m = mass of the person being lifted = 80 kg

Force equation for the motion of person in upward direction is given as

T - mg = ma

T = m (g + a)

T = (80) (9.8 + 0.70)

T = 840 N

b)

d = distance traveled in upward direction = 13 m

`W_(t)` = Work done by tension force

Work done by tension force is given as

`W_(t)` = T d

`W_(t)` = (840) (13)

`W_(t)` = 10920 J

c)

d = distance traveled in upward direction = 13 m

`W_(g)` = Work done by person's weight

Work done by person's weight is given as

`W_(g)` = - mg d

`W_(g)` = - (80 x 9.8) (13)

`W_(g)` = - 10192 J

d)

`F_(net)` = Net force on the person = ma = 80 x 0.70 = 56 N

v₀ = initial speed of the person = 0 m/s

v = final speed

Using work-energy theorem

`F_(net)` d = (0.5) m (v² - v₀²)

(56) (13) = (0.5) (80) (v² - 0²)

v = 4.3 m/s