Assuming this is a continuation of an earlier question [20398149], the crate starts at rest and is allowed to slide down the ramp with acceleration a = 2.35 m/s², so the block will have some speed as it reaches the bottom of the ramp and it will continue to slide some distance along the flat part.
If the flat part is made of the same material as the ramp, then in addition to its own weight and the normal force, the crate will also feel some friction that slows down its leftward slide.
• The crate's weight and the normal force act vertically, so that
∑ F = n - w = 0
n = w = mg = (100 kg) (9.8 m/s²) = 980 N
(where ∑ F = net force, n = magnitude of normal force, w = crate weight, m = crate mass, g = mag. of gravitational acceleration)
• The friction acts horizontally, so
∑ F = -f = m a
(where f = mag. of friction and a = crate acceleration)
The surface has a coefficient of kinetic friction of µ = 0.3, so
f = µ n = 0.3 (980 N) = 294 N
So at the bottom of the ramp, there are 3 forces exerted on the crate:
• its weight of 980 N pointing downward
• the normal force of the surface pushing upward on the crate, also of 980 N
• friction of 294 N, pointing to the right
and the two vertical forces cancel each other.
