Kinetic Friction And Deceleration

Question

A hockey puck sliding across rough ice with an initial speed of 12 m/s comes to rest after travelling some distance, the coefficient of kinetic friction being 0.3. What is the deceleration of the puck (take g = 10 m/s²)?

RankGuru · Accepted Answer

As outlined in NCERT Class 11, Chapter 5 (Laws of Motion), kinetic friction acts on a moving body opposite to its motion and has magnitude f = μ_k N. On flat ice the normal reaction equals the weight, so N = mg, giving f = μ_k mg. By Newton's Second Law, the deceleration is a = f / m = μ_k g, which is independent of the puck's mass. Substituting, a = 0.3 × 10 = 3 m/s². The puck slows steadily until it stops. Option 0.3 m/s² mistakenly forgets to multiply by g and just reports μ_k. Option 30 m/s² multiplies μ_k by g and then by an extra factor of 10, an order-of-magnitude slip. Option 1.5 m/s² halves the correct value without basis. A notable consequence of the mass cancelling is that a heavy puck and a light puck launched at the same speed on the same ice would slide for the same distance before stopping, since both decelerate identically. This counter-intuitive result is a direct signature of friction being proportional to the normal force, which itself scales with mass. Plausibility check: the deceleration must be only a fraction of g for a small friction coefficient, and 3 m/s² is exactly 0.3 g, which is consistent and correctly expressed in m/s², confirming the answer.

Kinetic Friction And Deceleration

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