help with kinetics pleaseeeeeeeeeee!!!!?

Question

a 3.0 kg block starts from rest at the top of a 30 degree angle incline and accelerates uniformly down the incline, moving 2.00 m in 1.50s .

it asks for accelertation which i got and its= 1.78 m/s^2
asks for speed of block after it has slid a distance of 2.0 which i got and its = 2.67 m/s
it asks for frictional force on the block and coefficient of kinetic energy between the block and incline i cant get these two can some one help me PLEASE!!!!!!!!!!!

Answer 1

If the block is on a 30 deg. incline, gravity acts vertically. A portion of the force of gravity (3 kg * 9.8 m/s^2 = 29.4 N) will act perpendicular to the inclined plane (this portion will push the block against the surface of the plane, and cause friction) and one portion will be parallel to the inclined plane (responsible for the block sliding downhill).

The downhill portion of the force is 29.4 * Sin(30) = 14.7 N

If the totality of this force was used to accelerate the block, it should accelerate at:
F = m a
a = F/m = 14.7 / 3 = 4.9 m/s^2

But the block is only accelerating at 1.78 m/s^2. Obviously, friction is slowing down. How much friction?

Let's figure out how much of the force is used for acceleration:

F = m a = 3 * 1.78 = 5.34 N

Therefore, (14.7 - 5.34 =) 9.36 Newtons were used to fight friction (the frictional force)
(We are assuming that the coefficient of friction remained constant during the acceleration -- that is a reasonnable assumption in most cases).

The friction is the product of the coefficient of friction (let's call it k) times the force that pushes the block against the surface of the plane.

9.36 = k * (29.4*Cos(30) ) = k*25.46

Therefore, k = 9.36 / 25.46 = 0.368 (coefficient of friction).

(I have assumed that your 1.78 m/s^2 is correct -- I did not check it)