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

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

Step-by-step 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