Halfway done Force/Friction Problem

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A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic friction between the block and the plane is µk = 0.10. The block is attached to a second block of mass m2 = 220 g that hangs freely by a string that passes over a frictionless and massless pulley.
Find its speed when the second block has fallen 30.0 cm.
cm/s

What I have done:
I turned the 2 masses into kg.

M1 = .29kg and M2 = .22kg
M1 Force of Gravity = .29kg * 9.81 m/s^2 = 2.8N
M1 Normal Force = 28 * cos30 = 2.4N
M1 Parallel Force = 28 * sin30 = 1.4N
M1 Friction Force = .10 * 2.4 = .24N

M2 Force of Gravity = 2.1N

This where I am stuck, I don't know what I should do next to determine the speed of M2 when it is falling.Edit:

The Net Force would be
1.4N - .24N = 1.16N
= 2.1N - 1.16N = .94N

F = M * A
.94N = .51kg * A
A = 1.8 m/s^2

Since there is 100 centimeters in a meter
so A = 180 cm/S^2

Speed = Distance * Time
X = 30cm * T
Where T = 30/180 = .17s
30 * .17s
Speed = 5.1cm/s

Is this correct?

 

[edziura]

The Net Force would be
1.4N - .24N = 1.16N
= 2.1N - 1.16N = .94N

The friction force on m1 acts opposite the direction of motion of m1 with respect to the inclined plane and so it acts down the plane.

Once a is found, you can use v = a t and d = (1/2) a t^2 to find v.

 

[ff4930]

I found
F = M * A
.94N = .51kg * A
A = 1.8 m/s^2

Since there is 100 centimeters in a meter
so A = 180 cm/S^2

I found Time to be .58s
V = to be 104cm/s

Is this correct?

 

[ff4930]

Can someone verify if this is the correct solution? Thanks!

 

[ff4930]

Anyone? I just need to verify this solution before I submit it online, please.