A2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) what is the block’s acceleration down the ramp and the force of the ramp on the block? (b) what force applied upward along and parallel to the ramp would allow the block to move with constant velocity?

The block’s acceleration down the ramp is 4.9m/s²

The force applied upward along and parallel to the ramp would allow the block to move with a constant velocity is 9.8 N

According to Newton's second law;

Fm is the moving force

m is the mass of the block

a is the acceleration

Since the block acts along the ramp

Given the following parameters

m = 2.0kg

g = 9.8m/s²

= 30 degrees

Get the required acceleration:

Substitute the given parameters

Hence the block’s acceleration down the ramp is 4.9m/s²

b) The force applied upward along and parallel to the ramp is the moving force along the plane expressed as:

Hence the force applied upward along and parallel to the ramp would allow the block to move with a constant velocity is 9.8 N

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a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N