A hockey player weighing 70 kg, standing on the ice, throws a puck weighing 0.3 kg in the horizontal direction

| January 11, 2021 | education

A hockey player weighing 70 kg, standing on the ice, throws a puck weighing 0.3 kg in the horizontal direction at a speed of 10 m / s. How far will a hockey player roll if the friction force acting between him and the ice is 14N.

To determine the distance that the hockey player should roll back, we use the formula: S = Vx ^ 2 / 2a = (Vsh * msh / mx) ^ 2 / (2 * Ftr / mx) = Vsh ^ 2 * msh2 / (2 * Ftr * mx).
Values of variables: Vsh – throwing speed of the puck (Vsh = 10 m / s); msh is the mass of the washer (msh = 0.3 kg); mx is the mass of the hockey player (mx = 70 kg); Ftr is the effective friction force (Ftr = 14 N).
Calculation: S = Vsh ^ 2 * msh ^ 2 / (2 * Ftr * mx) = 10 ^ 2 * 0.3 ^ 2 / (2 * 14 * 70) = 0.00459 m or 4.59 mm.
Answer: The hockey player must roll back 4.59 mm.



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