Posted by k9 on .
There is a block on top of an incline plane. The incline is going down to the right. The angle to the right of the 90 is 36.9 degree. There is a force pushing on it from the left.
Find the applied force (F) needed to prevent the block from sliding down the incline; ignore friction. The mass of the block is .5 kg and the mass of the incline plane is 2kg.
Part 2 What is the acceleration of the system?
I have the block as .5 kg. mg = 4.9N Fn = cos 36.9 degrees x 4.9 x .5= 3.92 N
mg sin 36.9 degrees = 2.94 N
I believe acceleration is 5.88 m/s 2 for the block
I am stuck.
1. Fb = mg = 0.5kg * 9.8N/kg = 4.9N = Force or weight of block in Newton.
Fp = 4.9sin36.9 = 2.94N = Component
parallel to the plane.
We don't need the comp. perpendicular to the plane, because there is no friction.
To prevent the block from sliding down the plane, the net force(Fn) must be 0:
Fn = Fap - Fp = 0,
Fap = Fp = 2.94N Parallel to the plane
and pulling up the plane.
2. Since the net force(Fn) is zero, there is no acceleration:
F = ma,
a = Fn/m = 0/0.5kg = 0 m/s^2.