Answer:
[tex]\mu_s = 0.62[/tex]
[tex]\mu_k = 0.415[/tex]
The motion of the block is downwards with acceleration 1.7 m/s^2.
Explanation:
First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.
[tex]x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2[/tex]
Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.
[tex]F = ma\\F = 1.7m[/tex]
Now, let’s investigate the free-body diagram of the block.
Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,
[tex]F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415[/tex]
As for the static friction, we will consider the angle 31.8, but just before the block starts the move.
[tex]mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62[/tex]
