A hockey puck is given an initial speed of
19 m/s on a frozen pond. The puck remains
on the ice and slides 113 m before coming to
rest.
The acceleration of gravity is 9.8 m/s
2
.
What is the coefficient of friction between
the puck and the ice?
(1/2) M Vo^2 = mu*M*g*X
= (work done to stop)
M cancels out
mu = Vo^2/(2*g*X) = 0.163
is the kinetic friction coefficient
To find the coefficient of friction between the puck and the ice, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity of the puck (0 m/s since it comes to rest)
- vi is the initial velocity of the puck (19 m/s)
- a is the acceleration of the puck (which we want to find)
- d is the distance covered by the puck (113 m)
Rearranging the equation, we have:
a = (vf^2 - vi^2) / (2d)
Since vf is 0 m/s, we can simplify it further:
a = -vi^2 / (2d)
Now, we can calculate:
a = -(19 m/s)^2 / (2 * 113 m)
a = -361 m^2/s^2 / 226 m
a ≈ -1.598 m/s^2
Now, we know that the frictional force acting on the puck is given by:
Ff = μN
Where:
- Ff is the frictional force
- μ is the coefficient of friction (which we want to find)
- N is the normal force (equal to the weight of the puck)
The normal force can be calculated by multiplying the mass of the puck by the acceleration due to gravity:
N = m * g
Where:
- m is the mass of the puck (which is not given in the given information)
- g is the acceleration due to gravity (9.8 m/s^2)
As the mass of the puck is not given, we can cancel it out in our calculation for the coefficient of friction. Thus, we have:
Ff = μ * g
Since the puck is in equilibrium, the frictional force is equal to the gravitational force:
Ff = m * g
Therefore, we can conclude that:
μ * g = m * g
This implies that the coefficient of friction (μ) is equal to 1.
So, the coefficient of friction between the puck and the ice is 1.