A hockey puck struck by a hockey stick is given an initial speed v0 in the positive x-direction. The coefficient of kinetic friction between the ice and the puck is μk.
(a) Obtain an expression for the acceleration of the puck. (Use the following as necessary: μk and g.)
(b) Use the result of part (a) to obtain an expression for the distance d the puck slides. The answer should be in terms of the variables v0, μk, and g only.
a) The acceleration is caused by the friction, which is μkFn. It is the only force acting in the x-direction, and therefore Newton's second law for this situation is: μkFn = -ma. (The acceleration is negative because it is in the opposite direction of the motion, which is defined in the problem as positive.) Fn is mg, assuming the ice is level. So: μkmg = -ma. The mass cancels and a = -μkg. b) The distance the puck moves can be found using the kinematic equation: v^2 = vi^2 + 2a(x - xo) Note that d = x - xo, and v = 0 (the puck comes to rest). So the equation becomes 0 = vi^2 + 2ad, and d = -vi^2/2a. Since a = -μkg: d = (vi^2)/(2μkg)
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