A stone sliding on the ice surface stops 10 seconds after the start of movement.

A stone sliding on the ice surface stops 10 seconds after the start of movement. Determine the initial speed of the stone if the coefficient of friction is 0.05

t = 10 s.

V = 0 m / s.

g = 9.8 m / s ^ 2.

μ = 0.05.

V0 -?

Let us write Newton’s 2 law in vector form: m * a = m * g + Ffr + N, where m is the body mass, a is the acceleration of the body, g is the acceleration of gravity, Ffr is the friction force, N is the ice reaction force.

ОХ: m * a = Ftr.

OU: 0 = m * g – N.

Ftr = μ * N = μ * m * g.

m * a = μ * m * g.

The body moves with acceleration a, which is determined by the formula: a = μ * g.

a = (V – V0) / t, since V = 0 m / s, the body has stopped, then the formula will take the form: a = – V0 / t.

The sign “-” means that the acceleration is directed in the opposite direction to the movement.

μ * g = V0 / t.

V0 = μ * g * t.

V0 = 0.05 * 9.8 m / s ^ 2 * 10 s = 4.9 m / s.

Answer: the initial velocity of the stone was V0 = 4.9 m / s.