A 0.5 kg ball rolling freely on a horizontal surface has decreased its speed

A 0.5 kg ball rolling freely on a horizontal surface has decreased its speed from 10 m / s to 4 m / s. What is the work of the friction force equal to?

Data: m (mass of a rolling ball) = 0.5 kg; V1 (initial ball speed) = 10 m / s; V2 (final speed) = 4 m / s; we do not take into account the kinetic energy of the ball’s rotational motion (Ekw = I * ω2 / 2 ≈ 0 m).

To calculate the required work on the ball of the friction force, we apply the formula: Atr = ΔEk = m * (V1 ^ 2 – V2 ^ 2) / 2.

Calculation: Atr = 0.5 * (10 ^ 2 – 4 ^ 2) / 2 = 21 J.

Answer: The frictional force performed a work on the ball of 21 J.