For part (ii) your answer would be correct if there is no friction, but there is.

I suggest you start by drawing a free body diagram for both bodies. For the 2Kg body you have forces of weight acting down and tension in the rope acting upward. For the 5 Kg mass you have tension pulling to the right and friction pulling to the left. Because both bodies are tied by a rope the magnitude of the tension acting on both bodies is equal, and the acceleration of both bodies is equal. You can set up an equation that equates the acceleration of both bodies - this will allow you to solve for T. Once you have that, you can solve for the acceleration.

Ah ok. I see where I went wrong.

ii) T - 4.9 N = 5 kg (a)

-T + 0.9800 N = 2 kg (a)

--------------------------------

-3.92 N = 7 kg (a)

-0.56 m/s^2 = a

T = 5 kg (-0.56 m/s^2) + 4.9 N = 2.1 N

That should be right now.

i) As for the first one, is it possible someone can give me some hints? I truly and honestly don't know what I'm doing. This is as far as I've gotten:

total mass = 7 kg

vf = 0 m/s

vi = ?

a = -0.56 m/s^2

Kinetic friction = 4.9 N

I don't have a formula I can think of to 100% use

I don't know distance or time, so the Kinematics formulas don't look like they'll help me. I am absolutely stuck.

All I could muster was this:

Kinetic Energy = Kinetic Friction

Thus v = sqrt( Kinetic Fricition / .5(mass))

v = sqrt( 4.9 N / .5(7))

v = 3.74 m/s

Thanks for the help so far!