Friction and Velocity

[mmiller39]

Here is a problem I am having difficulty with:

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.89 m/s. However, this speed is inadequate to compensate for the kinetic friction 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 reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?


What we know:

Vo = 2.89 m/s <---- this speed only derives half of the expected result
fk angle is Cos 180 = -1, which is the direction of the force acting on the puck.

I am completely confused as to where to begin here.

Any help would be appreciated.

Thanks,

Matt

[OlderDan]

Here is a problem I am having difficulty with:

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.89 m/s. However, this speed is inadequate to compensate for the kinetic friction 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 reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?


What we know:

Vo = 2.89 m/s <---- this speed only derives half of the expected result
fk angle is Cos 180 = -1, which is the direction of the force acting on the puck.

I am completely confused as to where to begin here.

Any help would be appreciated.

Thanks,

Matt
The problem is telling you that friction causes a certain (but unknown) deceleration of the puck. It is also telling you that the puck needs enough initial velocity to travel twice as far as it did with a known initial velocity. What equation do you know that relates velocity changes to acceleration and distance travelled?

[mmiller39]

Thanks for the response,

I am thinking that I would use the Kinematics in one dimension type equation for this:

V^2 = Vo^2 + 2ax

Which would lead me to believe that the initial velocity should be the square of the original initial velocity that only made it half way.

-Matt

[OlderDan]

Thanks for the response,

I am thinking that I would use the Kinematics in one dimension type equation for this:

V^2 = Vo^2 + 2ax

Which would lead me to believe that the initial velocity should be the square of the original initial velocity that only made it half way.

-Matt
Not quite the right conclusion, but it is the right equation. In both cases you have a negative acceleration and zero final velocity. In going from the first case to the second case you are trying to double the x. What do you have to do to Vo to double the x?