A stone of 1kg is thrown with a velocity of 20ms⁻¹ across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?

Before we proceed we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force is acting in the leftward direction while the stone is moving in the rightward direction. One thing we have to fix in our mind is that frictional force will always act opposite to the direction of motion of the body. Now, on resolving the forces, f = ma ———(1) Also, from the equation of motion v² = u² + 2as, here final velocity is zero, so v = 0 u² = -2as a = -u²/2s = -20²/2 × 50 = -4 m/sec² On substituting the given data in equation (1) f = 1 × (-4) = -4N