A box and its contents have a total mass M = 9 kg. A string passes through a hole in the box, and you pull on the string with a constant force F = 77 N (this is in outer space; there are no other forces acting).

Question

(a) Initially the speed of the box was vi = 5 m/s. After the box had moved a distance w = 2.1 m, your hand had moved an additional distance d = 0.4 (a total distance of w + d), because additional string of length d came out of the box (see figure). What is now the speed vf of the box?

Answer 1

you do
Fd=.5m*vfinal^2-.5m*vinitial^2
solve for vfinal and that is the correct answer.

Answer 2

You wasted one of my tries @ jm

Answer 3

F*(w+d)= 1/2*m*Vfinal - 1/2*m*Vinitial

and just solve for vfinal

Vfinal = squaredroot/((F(w+d) + .5*m*Vinitial)/ (.5*m))

Answer 4

The correct way to find the solution to this problem is as follows:
F*w=(0.5*m*v(final)^2) - (0.5*m*v(initial)^2)
So for this problem specifically,
77*2.1 = (0.5 * 9 * vf^2) - (0.5 * 9 * 5^2)
161.7 = (4.5 * vf^2) - 112.5
274.2 = 4.5* vf^2
60.93 = vf^2
v(final) = sqrt(60.93)
v(final) = 7.81 m/s