Question

3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration (in m/s2) down the ramp? (Enter the magnitude.) m/s2 What is the force (in N) of the ramp on the block? (Enter the magnitude.) N (b) What force (in N) applied upward along and parallel to the ramp would allow the block to move with constant velocity?

Answer 1

Answer:

a)  a = 7.72 m / s²,  N = 19.9 N  and b)   F = 25.5 N

Explanation:

To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

    sin52 = Wx / W

    cos52 = Wy / W

    Wx = W sin52

    Wy = w cos 52

Let's write them equations

X axis

    Wx = ma

Y Axis

    N-Wy = 0

    N = Wy

a) Let's calculate the acceleration

    a = W sin52 / m = mg sin 52 / m

    a = g sin 52

    a = 9.8 sin52

    a = 7.72 m / s²

The force of the ramp is normal

    N = Wy = mg cos 52

    N = 3.3 9.8 cos 52

    N = 19.9 N

b) For the block to move at constant speed the sum of force on the axis must be zero,

    F - Wx = 0

    F = Wx

    F = mg sin52

   F = 3.3 9.8 sin 52

   F = 25.5 N

Parallel to the plane and going up