A boy weighing 50 kg, having rolled down the hill on a sled, drove to a stop 20 m in 10 seconds.

A boy weighing 50 kg, having rolled down the hill on a sled, drove to a stop 20 m in 10 seconds. What is the friction force equal to?

m = 50 kg.
S = 20 m.
t = 10 s.
V = 0 m / s.
Ftr -?
Three forces act on the boy: the force of gravity m * g vertically downward, the reaction force of the support N along which the sled goes, directed vertically upward, and the friction force Ffr, which is directed in the opposite direction of movement.
The force of gravity and the reaction force of the support N cancel each other out.
Horizontally the sled goes under the action of one frictional force Ftr.
Let’s write 2 Newton’s law: m * a = Ftr.
Let us find the acceleration of the body a by the formula of uniformly accelerated motion:
S = (V ^ 2 – V0 ^ 2) / 2 * a, since V = 0 m / s, then S = – V0 ^ 2/2 * a.
a = (V – V0) / t, since V = 0 m / s, then a = – V0 / t.
S = – V0 ^ 2 * t / 2 * (- V0) = V0 * t / 2.
V0 = 2 * S / t.
a = 2 * S / t ^ 2.
Ftr = 2 * S * m / t ^ 2.
Ftr = 2 * 20 m * 50 kg / (10 s) ^ 2 = 20 N.
Answer: friction force Ffr = 20 N.