Kinetic Friction of hockey puck

[nambod]

Homework Statement

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.71 m/s. However, this speed is inadequate to compensate for the kinetic friction that exists between the puck and the ice. As a result, the puck travels only one-half the distance between the players before sliding to a halt. What minimum initial speed should the puck have been given so that it just reached the teammate? Assume the force of kinetic friction is the same on the puck everywhere between the two players. Also assume the ice surface is flat and μk = 0.1.

Homework Equations

The Attempt at a Solution

 

[Matterwave]

Have you even tried doing this problem?

Can you set up the necessary equations? Do you want to do this from an energy or force perspective? Either way works.

 

[kerimek]

I would approach this problem by first understanding the concept of kinetic friction. Kinetic friction is the force that opposes the motion of an object as it slides over a surface. It is dependent on the coefficient of kinetic friction (μk) and the normal force (N) between the two surfaces. In this case, the normal force is equal to the weight of the puck, which we can calculate using its mass (m) and the acceleration due to gravity (g).

Next, I would use the equation Fk = μkN to calculate the force of kinetic friction acting on the puck. Since the force of kinetic friction is the same throughout the distance between the players, we can set it equal to the net force acting on the puck in the direction of motion. This net force is equal to the mass of the puck (m) multiplied by its acceleration (a). We can rearrange this equation to solve for the acceleration, which is the rate of change of velocity.

a = μkN/m

Now, we can use the equation for average velocity (v = u + at) to calculate the minimum initial speed (u) needed for the puck to reach the teammate. In this case, the final velocity (v) is 0, since the puck comes to a halt. The distance between the players is given as one-half of the total distance, so we can use this value for the displacement (s).

0 = u + (μkN/m)t
u = - (μkN/m)t

Substituting the values for μk, N, m, and t given in the problem, we get:

u = - (0.1)(mg)/m(t)
u = - 0.1gt

Since we want the puck to reach the teammate, we can set the final displacement (s) equal to one-half of the total distance and solve for t.

s = (1/2)ut
(1/2)d = (1/2)(-0.1gt)t
d = - 0.05gt^2

Finally, we can use the equation for displacement (s = ut + (1/2)at^2) to solve for the minimum initial speed (u).

d = ut + (1/2)at^2
- 0.05gt^2 = ut + (1/2)(μkN/m)t^2
u =