I’ve never been curling. Even so, I can imagine the following question resonates with many curlers and spectators: Solution:<\/p>\n From my previous post<\/a>, the force of friction is calculated by<\/p>\n Ff<\/sub> = μFN<\/sub><\/p>\n In this case, the stone is on level ice, so<\/p>\n Ff<\/sub> = μFN<\/sub> = μmg<\/p>\n The force of friction is the only unbalanced force acting on the stone, so its acceleration is due only to friction. From Newton’s Second Law,<\/p>\n a = F\/m = μmg\/m = μg = 0.010(9.8) = 0.098m\/s2<\/sup><\/p>\n Since the acceleration opposes the velocity, its sign is negative:<\/p>\n a = -0.098m\/s2<\/sup><\/p>\n Now, we can calculate the distance the stone travels using kinematics:<\/p>\n v2<\/sub>2<\/sup> = v1<\/sub>2<\/sup> + 2ad<\/p>\n where v1<\/sub>= initial velocity = 1.2m\/s, v2<\/sub> = 0m\/s<\/p>\n Therefore,<\/p>\n 02<\/sup>=1.22<\/sup> + 2(-0.098)d<\/p>\n 0 = 1.44 – 0.196d<\/p>\n We add 0.196d to both sides:<\/p>\n 0.196d = 1.44<\/p>\n We divide both sides by 0.196:<\/p>\n d = 1.44\/0.196 = 7.3469…m (7.3m in sig figs)<\/p>\n Apparently the stone will glide 7.3m before stopping.<\/p>\n Source:<\/p>\n Heath, Robert W et al. Fundamentals of Physics<\/u>. D.C. Heath Canada Ltd., 1981.<\/p>\n
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\nA curling stone is released at 1.2m\/s. If the coefficient of friction between the ice and the stone is 0.010, how far will it travel before coming to rest?<\/strong><\/p>\n