The sleds, sliding on a horizontal surface, stopped after passing 25 m. Determine the initial

The sleds, sliding on a horizontal surface, stopped after passing 25 m. Determine the initial speed of the sled if the friction coefficient is 0.05.

S = 25 m.

V = 0 m / s.

g = 10 m / s2.

μ = 0.05.

V0 -?

Since the sled on a horizontal surface moved uniformly accelerated to a complete stop, then their path S is expressed by the formula: S = V02 / 2 * a, where a is the acceleration during braking.

V0 = √ (2 * S * a).

When braking sleigh 2, Newton’s law has the form: m * a = Ftr + m * g + N, where m * g is the force of gravity, N is the reaction force of the surface of the inclined plane, Ftr is the friction force.

ОХ: m * a = Ftr.

OU: 0 = N – m * g.

N = m * g.

The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g.

m * a = μ * m * g.

a = μ * g.

V0 = √ (2 * S * μ * g).

V0 = √ (2 * 25 m * 0.05 * 10 m / s2) = 5 m / s.

Answer: the sleigh had an initial speed of V0 = 5 m / s.