Answer:


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.

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

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,

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

Answer:
Final Velocity = 4.9 m/s
Explanation:
We are given;. Initial velocity; u = 2 m/s
Constant Acceleration; a = 0.1 m/s²
Distance; s = 100 m
To find the final velocity(v), we will use one of Newton's equations of motion;
v² = u² + 2as
Plugging in the relevant values to give;
v² = 2² + 2(0.1 × 100)
v² = 4 + 20
v² = 24
v = √24
v = 4.9 m/s
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