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 =
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)
(a) To obtain the expression for the acceleration of the puck, we need to consider the forces acting on the puck. The only horizontal force acting on the puck is the force of kinetic friction.
The force of kinetic friction is given by the equation: f(kinetic) = μk * N
Where μk is the coefficient of kinetic friction and N is the normal force acting on the puck. In this case, N is equal to the weight of the puck since it is on a horizontal surface. The weight of the puck is equal to m * g, where m is the mass of the puck and g is the acceleration due to gravity.
The net force acting on the puck can be calculated using Newton's second law:
Fnet = m * a
Since the force of kinetic friction is the only horizontal force, it is equal to the net force acting on the puck:
f(kinetic) = Fnet
μk * N = m * a
Substituting N = m * g, we get:
μk * m * g = m * a
Simplifying, we find:
a = μk * g
So, the expression for the acceleration of the puck is a = μk * g.
(b) To obtain the expression for the distance the puck slides, we can use the kinematic equation:
v^2 = v0^2 + 2 * a * d
Where v is the final velocity of the puck, v0 is the initial velocity of the puck, a is the acceleration of the puck (which we found in part a), and d is the distance the puck slides.
Since the puck eventually comes to a stop, the final velocity v is zero. Thus, the equation simplifies to:
0 = v0^2 + 2 * a * d
Solving for d, we get:
d = -v0^2 / (2 * a)
Substituting the expression for acceleration we found in part a, we have:
d = -v0^2 / (2 * μk * g)
So, the expression for the distance the puck slides is d = -v0^2 / (2 * μk * g). Note that the negative sign indicates that the distance is measured in the opposite direction of the initial velocity.