A 200kg car is coasting to the right and slowing down due to a friction force of 200N.

By Newton's second law, the net force on the car is

∑ F = m a

where m is the mass of the car and a is magnitude of the acceleration.

The car is slowing down, so the net force opposes the direction in which the car is moving. Vertically, the car is in equilibrium, the net force is entirely due to friction. Taking the direction opposite the car's motion to be negative, we have

- 200 N = (200 kg) (-a)

a = (200 kg) / (200 N)

a = 1 m/s²

If you're also supposed to find the coefficient of kinetic friction, use the fact that the car the vertical forces cancel out, namely

n + (-w) = 0

where n is the magnitude of the normal force and w is the car's weight. The friction force f is proportional to the normal force n by a factor of µ, the coefficient of kinetic friction, such that f = µ n. Then

f / µ = w

(200 N) / µ = (200 kg) (9.80 m/s²)

µ = (200 N) / ((200 kg) (9.80 m/s²))

µ ≈ 0.102

real, smaller and upside-down.   so the answer should be b.