A skater weighing 60 kg, standing on skates on the ice, throws an object weighing 1 kg

A skater weighing 60 kg, standing on skates on the ice, throws an object weighing 1 kg in a horizontal direction at a speed of 24 m / s and rolls back 40 cm. Find the coefficient of friction of skates on the ice.

Given:

m1 = 60 kilograms – the mass of the skater;

m2 = 1 kilogram – the weight of the load thrown by the skater;

g = 10 m / s ^ 2 – acceleration of gravity;

v2 = 24 m / s – the speed with which the skater threw a stone.

S = 40 centimeters = 0.4 meters – the path that the skater traveled after throwing a stone.

It is required to determine the coefficient of friction of the skater on ice k.

Let’s find the speed that the skater got after throwing a stone:

m1 * v1 = m2 * v2,

v1 = m2 * v2 / m1 = 1 * 24/60 = 0.4 m / s.

Let’s find the acceleration that the skater acquired:

S = v1 * t – a * t ^ 2/2 = v1 ^ 2 / a – v1 ^ 2 / (2 * a) = v1 ^ 2 / (2 * a), hence:

a = v1 ^ 2 / (2 * S) = 0.4 ^ 2 / (2 * 0.4) = 0.4 / 2 = 0.2 m / s ^ 2.

Then the coefficient of friction will be:

k * g = a;

k = a / g = 0.2 / 10 = 0.02.

Answer: The coefficient of friction is 0.02.