What Is the Coefficient of Kinetic Friction for a Hockey Puck on Ice?

  • Thread starter Thread starter bbreezy
  • Start date Start date
  • Tags Tags
    Friction
AI Thread Summary
To calculate the coefficient of kinetic friction for a hockey puck sliding on ice, the initial speed of 14.1 m/s and stopping distance of 198.9 m are used. The acceleration can be determined using the kinematic equation vi² + 2ad = 0, which helps find the necessary deceleration. Once the acceleration is calculated, it can be related to the net force acting on the puck, which is the frictional force. The correct formula for the coefficient of kinetic friction is indeed u_k = a/g, where 'a' is the acceleration and 'g' is the acceleration due to gravity. Accurate calculations are crucial for obtaining the correct value of the coefficient.
bbreezy
Messages
7
Reaction score
0
I am having trouble with this problem, it states:
A hockey puck on a frozen pond with an initial speed of 14.1 m/s stops after sliding a distance of 198.9 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.

I tried using the equation uk=a/g but didnt work out for me. Any help?
 
Physics news on Phys.org
Welcome to PF bbreezy!

Using basic kinematics, you can figure out what the object's acceleration must have been for it to require that stopping distance in order to come to rest from that initial speed.

Once you know the acceleration of the object, you know the net force on the object (due to Newton's second law). In this case, the net force is the frictional force, since it is the only one that acts on the object. So, equating the net force to the frictional force, you will indeed find that the expression you posted for the coefficient of friction is correct (since it will just turn out to be the ratio of the net force to the weight, and the m's will cancel).

So, if you didn't get the right answer, you must not have computed 'a' correctly. Can you post your calculations here?
 
Vi+2ad = 0
14.1+2(a)(198) = 0
2(a)(198) = -14.1

then this is where I think I am messing up
 
bbreezy said:
Vi+2ad = 0
14.1+2(a)(198) = 0
2(a)(198) = -14.1

then this is where I think I am messing up

It should really be:

vi2 + 2ad = 0
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top