here is the question:

a package has a mass of 40 kg and the coeficient of kinetic friciton between the package and the ground is 0.42. the parachute that is attached to it by a string has a mass of 10 kg. at the moment, the air is exerting a force of 200N to the right on the chute. what is the tension in the rope connecting the package to the parachute?

. ------->F air
Box-------------parachute

(tried to draw the diagram for u guys)

what is did was drew free body diagrams for both the chute and the box
ended up with the equation
F-fk=(m1+m2)a
solved for a

then subbed my accelration into my FBD equation of only the box so...
T-fk=m1*a
solved for T...
didnt get the right answer

how do u do this question?

up

Is the answer either 192 or 104 N?

answer is 129

i got an answer of 192 aswell..

Maybe it was a typo and meant 192 instead?

that would be pretty bad

im looking at my friends old test. he guessed 129 right lol

I got 192.96 N

Assuming the rope doesn't stretch, then by kinematic constraints the displacement between the chute and the box must remain constant, so the velocity and acceleration is equal. a(chute) = a(box).
a(chute) = m(chute)/(F(air)-T) = 10/(200-T)
a(box) = m(box)/(T-F(friction)) = 40/(T-40*9.81*0.42)
a(chute) = a(box)
m(chute)/(F(air)-T) = m(box)/(T-F(friction))
...
solves to T = (800+40*9.81*0.42)/5
= 192.96 N