A hockey player hits a puck with his stick, giving the puck an initial speed of 5.8m/s .
Part A
If the puck slows uniformly and comes to rest in a distance of 20m , what is the coefficient of kinetic friction between the ice and the puck?
x = (v-u)(v+u)/2a
where x = distance, v = final speed, u = initial speed, a = acceleration
a = (v-u)(v+u)/2x = -5.8*5.8/40 = -0.841
f=ma
u(k) * f(normal) = ma
u(k) *m *g = ma
u(k) *g = a
u(k) = a/g
u(k) = -0.841 / 9.81 = 0.086
where u(k) = coefficient of kinetic friction
mv²/2 =W(fr) = μNs= μmgs
μ=v²/2gs =5.8²/2•9.8•20 =0.086
Well, if the puck comes to a stop, I guess we can call it a "puck brake" moment. It must have really wanted a break from all that sliding around on the ice!
Anyway, it seems like we need to find the coefficient of kinetic friction between the ice and the puck. Let's see what we can do.
First, we'll need to use the equation for uniform acceleration: v^2 = u^2 + 2as. Here, v is the final velocity (which is 0, since the puck comes to rest), u is the initial velocity (5.8 m/s), a is the acceleration, and s is the distance (20 m).
Rearranging the equation, we get a = (v^2 - u^2) / (2s).
Since the puck slows uniformly, we can assume a constant acceleration. And since the puck comes to rest, v is 0. Plugging in the values, we have a = (0^2 - 5.8^2) / (2 * 20).
Simplifying, we find a = -7.406 m/s^2.
Now, the force of friction can be calculated using the equation F = m * a, where m is the mass of the puck. However, we're trying to find the coefficient of kinetic friction, so we can use another equation: F = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force.
In this case, N is the weight of the puck, which can be calculated using the equation N = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since we're interested in μ_k, we can rearrange the equation to find μ_k = F / N.
Now, to get the force of friction, we can use the equation F = m * a. However, we don't have the mass of the puck. So, unfortunately, we can't find the coefficient of kinetic friction without more information.
Looks like this question is a real "slippery" one! Maybe we can try another one?
To find the coefficient of kinetic friction between the ice and the puck, we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the puck comes to rest)
u = initial velocity of the puck (5.8 m/s)
a = acceleration due to friction
s = distance covered by the puck (20 m)
Rearranging the equation to solve for the acceleration (a):
a = (v^2 - u^2) / (2s)
Plugging in the given values:
a = (0 - (5.8)^2) / (2 * 20)
Simplifying:
a = (-33.64) / 40
a = -0.841 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
Next, we need to find the frictional force acting on the puck. The frictional force (F) can be calculated using:
F = μN
where:
μ = coefficient of kinetic friction (what we're trying to find)
N = normal force
The normal force acting on the puck is equal to its weight, which is given by:
N = mg
where:
m = mass of the puck (assuming it to be constant)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Combining the two equations:
F = μmg
Now, the frictional force is also given by:
F = ma
Substituting the value of acceleration:
μmg = m * (-0.841)
Simplifying:
μ = -0.841g / g
μ = -0.841
The coefficient of kinetic friction between the ice and the puck is approximately -0.841.
Note: The negative sign indicates the direction of the frictional force, which is opposite to the motion of the puck.