A 4.04 kg block slides down a smooth, frictionless plane having an inclination of 30◦. The acceleration of gravity is 9.8 m/s^2. The acceleration of the plane is 4.9 What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.37 m down the incline?
Answer in units of N.

Explanation:

What a lot of words to solve such a simple problem! The perpendicular force is the one that is pushing straight down on the plane. There is no side to side movement here or friction acting on this dimension at all. Perpendicular force is the same as the weight of the block. That's it! Perpendicular force is also normal force which is the same thing as weight:

w = mg so

w = (4.04)(9.8) and

w = 4.0 × 10¹ N

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.