Asked by Anonymous
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.)
a =
(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.
d =
(a) Obtain an expression for the acceleration of the puck. (Use the following as necessary: μk and g.)
a =
(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.
d =
Answers
Answered by
Dan
f = ma
μk * f(normal) = m*a
μk *m*g = m*a
μk *g =a
so a = μk*g
this is defined to be a negative acceleration since the friction force works in the opposite direction to the movement of the puck
vFinal = v0 + at
so t = (vF-v0)/a
d = t*(vF+v0)/2
so d = (vF+v0)(vF-v0)/2a
substitute in our 'known' values: a = μk*g, and vF = 0 (since the puck has stopped sliding)
d = (0+v0)(0-v0)/ 2(μk*g)
d = v0^2 / (2μk*g)
(eliminate negatives as d is positive)
μk * f(normal) = m*a
μk *m*g = m*a
μk *g =a
so a = μk*g
this is defined to be a negative acceleration since the friction force works in the opposite direction to the movement of the puck
vFinal = v0 + at
so t = (vF-v0)/a
d = t*(vF+v0)/2
so d = (vF+v0)(vF-v0)/2a
substitute in our 'known' values: a = μk*g, and vF = 0 (since the puck has stopped sliding)
d = (0+v0)(0-v0)/ 2(μk*g)
d = v0^2 / (2μk*g)
(eliminate negatives as d is positive)
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