A hockey player weighing 80 kg evenly moves on the horizontal ice surface to find the friction
| January 17, 2021 | education
A hockey player weighing 80 kg evenly moves on the horizontal ice surface to find the friction force if the sliding friction coefficient is 0.015.
These tasks: m (the mass of the hockey player in question) = 80 kg; μ (coefficient of sliding friction) = 0.015.
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
To determine the friction force acting on the considered hockey player, we use the formula: Ftr. = μ * N = μ * m * g.
Let’s calculate: Ftr. = 0.015 * 80 * 10 = 12 N.
Answer: A friction force of 12 N. acts on the hockey player in question.
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