How to find the coefficient of kinetic friction

[Ushitha Dissanayake]

Homework Statement

The speed of a 4.0-N hockey puck, sliding across a level ice surface, decreases at the rate of 0.61 m/s2. The coefficient of kinetic friction between the puck and ice is:
A) 0.062
B) 0.25
C) 0.41
D) 0.62
E) 1.2

Homework Equations

I'm not too sure but i know that :
If the body begins to slide along the surface, the magnitude of the
frictional force rapidly decreases to a value fk given by
fk =ukFN
where uk is the coefficient of kinetic friction

The Attempt at a Solution

fk =ukFN
0.61=uk4
uk=0.61/4
uk=0.1525
But i was wrong

 

[gneill]

Hi Ushitha Dissanayake

Welcome to Physics Forums!

Don't be too quick to plug in numbers, It's easier to follow your work if you work symbolically as much as possible.

You can use the $x_2$ and $x^2$ icons in the edit panel header to create subscripts or superscripts for your text-mode equations. The $\Sigma$ icon gives you access to a menu of math symbols that you can also use.

fk =ukFN
0.61=uk4
uk=0.61/4
uk=0.1525

I don't understand the second line in the above. You seem to be equating an acceleration to a force? (It's hard to tell because there are no units associated with the numeric values). Is FN the normal force (weight of the puck)?

You know the weight (in Newtons) of the puck so you can find its mass, right? Since you're given the puck's acceleration, can you then use Newton's 2nd law to find the force acting on it?

 

[Ushitha Dissanayake]

Oh i see, i was actually just using the kinetic energy formula but i didn't properly realize that ƒk was the force of kinetic energy. Thanks for the help

 

[CWatters]

Kinetic energy doesn't have a force. This is a question about force due to friction and one of Newton's laws.

 

[kuruman]

In addition to all of the above comments, consider using the information that you have a 4.0 N puck as opposed to a puck with a different weight. You do not appear to have used that information in your attempted solution.