Question

A helicopter lifts a 60 kg astronaut 17 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. (a) How much work is done on the astronaut by the force from the helicopter? J (b) How much work is done on the astronaut by her weight? J (c) What is the kinetic energy? J (d) What is the speed of the astronaut just before she reaches the helicopter?

Answer 1

Answer:

a) W = 10995.6 J

b) W = - 9996 J

c) Kf = 999.6 J

d) v = 5.77 m/s

Explanation:

Given

m = 60 Kg

h = 17 m

a = g/10

g = 9.8 m/s²

a) We can apply Newton's 2nd Law as follows

∑Fy = m*a     ⇒     T - m*g = m*a     ⇒    T = (g + a)*m

where T is the force exerted by the cable

⇒    T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)

⇒    T = 646.8 N

then we use the equation

W = F*d = T*h = (646.8 N)*(17 m)

W = 10995.6 J

b) We use the formula

W = m*g*h     ⇒    W = (60 Kg)(9.8 m/s²)(-17 m)

⇒    W = - 9996 J

c) We have to obtain Wnet as follows

Wnet = W₁ + W₂ = 10995.6 J - 9996 J

⇒    Wnet = 999.6 J

then we apply the equation

Wnet = ΔK = Kf - Ki = Kf - 0 = Kf    

⇒  Kf = 999.6 J

d) Knowing that

K = 0.5*m*v²    ⇒    v = √(2*Kf / m)

⇒    v = √(2*999.6 J / 60 Kg)

⇒    v = 5.77 m/s

Answer 2

Answer:

Explanation:

mass =60kg d = 17m  a=g/10

(a) work done on the astronaut by the force from the helicopter = fd

but f =m(g+a)

 w= m( g+g/10)d

wt = 11/10 mgd

w =11/10 * 60 *9.8 * 17 = 10995.6J  = 1IKJ

(b) workdone  by her weight = -mgh

   = 60*9.8* 17 = -9996J

(C) Kinetic energy = wt + w

                             = (10995.6 - 9996)J = 999.6J

(d) Kinetic energy =1/2m$v^{2}$

hence velocity = $\sqrt{2ke/m}$ = 5.777m/s