Solve Sliding Hockey Puck Problem: Coeff. of Kinetic Friction

[kchurchi]

Homework Statement

A hockey puck is sliding across a frozen pond with an initial speed of 7.5 m/s. It comes to rest after sliding a distance of 23.7 m. What is the coefficient of kinetic friction between the puck and the ice?

v_0x = initial speed = 7.5 m/s
v_fx = final speed = 0 m/s
Δx = x distance traveled = 23.7 m
f_k,IP = force of kinetic friction from ice on puck
N_IP = normal force from ice on puck
W_EP = weight force from Earth on puck
m_P = mass of puck
g = acceleration due to gravity

Homework Equations

ƩF_net = m*a
ƩF_x = -f_k,IP = m*a_x
ƩF_y = N_IP - W_EP = m*a_y → a_y = 0 m/s^2 → N_IP = m_P*g

The Attempt at a Solution

I attempted this logic...

f_k,IP = μ_k*N_IP → (-m*a_x)/(m_P*g) = μ_k

But then I hit a wall when trying to find the x-acceleration. Help?

 

[voko]

You need the equation of distance traveled under constant acceleration (deceleration in this case).

 

[kchurchi]

I am not allowed to have the equations of constant acceleration to solve problems. I must derive all equations myself using calculus. How do I even know that the acceleration is constant?

 

[SHISHKABOB]

there's only one force acting on the puck in the x-direction: the frictional force

and it depends on two constant values: the coefficient of friction and the normal force on the puck

therefore, the force is constant, therefore the acceleration is constant

set up the equation of motion

[itex]\Sigma F = m\ddot{x} = F_{friction}[/itex]

and go from there

I'm assuming that since they want you to derive the equations yourself, that you know how to do a differential equation, right?

 

[Nickie]

To solve this sliding hockey puck problem, we can use the formula for kinetic friction, which states that the force of kinetic friction is equal to the coefficient of kinetic friction multiplied by the normal force. In this case, the normal force is equal to the weight force from Earth on the puck, since the puck is not moving in the y-direction. Therefore, we can rewrite the equation as fk,IP = μk*WEP.

Next, we can use the equation for net force, which states that the net force is equal to the mass of the object multiplied by its acceleration. In this case, we can use the x-direction since the puck is sliding horizontally. Therefore, we can rewrite the equation as ƩFx = -fk,IP = m*ax.

Finally, we can substitute the values given in the problem into the equations and solve for the coefficient of kinetic friction. Plugging in the values for the initial and final speeds, as well as the distance traveled, we can find the acceleration of the puck using the equation Δx = v0x*t + 1/2*a*t^2. Solving for t and plugging it back into the equation for acceleration, we get ax = -0.316 m/s^2.

Plugging in the values for the mass of the puck and the acceleration into the equation for net force, we get -fk,IP = (mP*-0.316 m/s^2). Plugging this into the equation for kinetic friction and solving for μk, we get a coefficient of kinetic friction of μk = 0.067. This means that the coefficient of kinetic friction between the puck and the ice is 0.067, which is a relatively low value, indicating a relatively smooth surface for the puck to slide on.