A hockey puck leaves a player's stick with a speed of 8.60 and slides 38.0 before coming to rest. Find the coefficient of friction between the puck and the ice.
with the speed of 8.60m/s and slides 38m
(1/2) M Vo^2 = (Friction force)*X
= M*g*mu*X
mu is the coefficient of kinetic friction.
mu = (1/2)Vo^2/(g X)
You need to say what the units of initial velocity (Vo) and distance (X) are. Just putting down numbers is not enough.
thanks i got the answer :)
Well, I must say, that puck must have had a real slippery time skating on that ice! Anyway, let's get to work. We can use the equation of motion to find the coefficient of friction. The equation is:
vf^2 = vi^2 + 2ad
Where vf is the final velocity (which is 0 since the puck comes to rest), vi is the initial velocity (8.60 m/s), a is the acceleration (which we'll assume is constant), and d is the distance traveled (38.0 m).
Plugging in the known values, we get:
0^2 = (8.60)^2 + 2a(38.0)
0 = 73.96 + 76a
Now, let's solve for a:
76a = -73.96
a = -0.97 m/s^2
Now, we can find the coefficient of friction using the equation:
a = μg
Where μ is the coefficient of friction and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
-0.97 = μ(9.8)
μ = -0.097
So, the coefficient of friction between the puck and the ice is approximately -0.097. Remember to keep the negative sign in mind, maybe it means the puck wanted to do the moonwalk on the ice!
To find the coefficient of friction between the puck and the ice, we can use the equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity of the puck (0 m/s since it comes to rest)
- u is the initial velocity of the puck (8.60 m/s)
- a is the acceleration of the puck (which is negative because the puck is slowing down)
- s is the distance the puck slides (38.0 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we have:
a = (0^2 - 8.60^2) / (2 * 38.0)
Simplifying further, we get:
a = -73.96 / 76
Now, the force of friction (F) acting on the puck can be calculated using Newton's second law:
F = ma
Where:
- F is the force of friction
- m is the mass of the puck (which we assume to be constant and cancels out in this calculation)
- a is the acceleration of the puck
The force of friction is also related to the normal force (N) between the puck and the ice by:
F = μN
Where:
- μ is the coefficient of friction
Since the mass cancels out in this calculation, we can rewrite the equation as:
a = μg
Where:
- g is the acceleration due to gravity
Rearranging the equation to solve for μ, we have:
μ = a / g
Substituting the value of acceleration due to gravity (approximately 9.8 m/s^2), we can calculate μ:
μ = -73.96 / (76 * 9.8)
Simplifying the calculation, we find:
μ ≈ -0.096
Therefore, the coefficient of friction between the puck and the ice is approximately -0.096.