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Q: Place the block at the top of a 30 degree ramp. What is the net force on the block when it begins to slide down the ramp? What is the acceleration of the block down the ramp? SHOW WORK/CALCULATIONS

First, let

f = mag. of static friction

n = mag. of normal force

w = m g = mag. of gravitational force (the weight of the block)

m = mass of the block

g = 9.80 m/s² (the mag. of the acceleration due to gravity)

a = acceleration of the block

It's the static friction that interests us at the moment "when it begins to slide", meaning the precise moment at which static friction is at its maximum. At that point, the friction has magnitude f such that

f = 0.5 n

The problem is much easier to work through if you split up the forces into components acting parallel and perpendicular to the ramp. By Newton's second law, we have

• the net force acting parallel to the ramp is

∑ F = - f + w sin(30°) = m a

• and the net force acting perpendicular to the ramp is

∑ F = n - w cos(30°) = 0

Hence the net force on the block as it begins to slide acts only in the parallel direction. Note that we take the positive parallel direction to be the one in which the block slides down the ramp (i.e. opposing the friction force), and the positive perpendicular direction to be the same as the normal force.

Solve the second equation for n :

n = m g cos(30°) = (100 kg) (9.80 m/s²) cos(30°) ≈ 849 N

Then

f = 0.5 (849 N) ≈ 424 N

and so the net force on the block is

∑ F ≈ - 424 N + (100 kg) (9.80 m/s²) sin(30°) ≈ 65.6 N

Next, you want to find the acceleration of the block as it is sliding down the ramp, during which time kinetic friction kicks in. We have the same equations as above, except now f = 0.3 n. So we still have n ≈ 849 N, which gives

f = 0.3 (849 N) ≈ 255 N

∑ F = - f + w sin(30°) = m a

→   - 255 N + (100 kg) (9.80 m/s²) sin(30°) ≈ (100 kg) a

→   a ≈ 2.35 m/s²

i believe the answer is a hope it and is correct