Problem 35 - College Physics

Question

\(\bullet$$\bullet\) A hockey puck leaves a player's stick with a speed of 9.9 \(\mathrm{m} / \mathrm{s}\) and slides 32.0 \(\mathrm{m}\) before coming to rest. Find the coef- ficient of friction between the puck and the ice.

Step-by-Step Solution

Answer

The coefficient of friction is 0.156.

Step 1: Identify Known Values

We know the initial speed of the hockey puck, \( v_i = 9.9 \, \mathrm{m/s} \), it slides a distance \( d = 32.0 \, \mathrm{m} \) and comes to rest, thus the final speed \( v_f = 0 \, \mathrm{m/s} \). We aim to find the coefficient of friction \( \mu \).

Step 2: Use Kinematic Equation

We use the kinematic equation: \[ v_f^2 = v_i^2 + 2ad \] where \( a \) is the acceleration (or deceleration in this case). Substitute to find \( a \): \( 0 = (9.9)^2 + 2a(32.0) \).

Step 3: Solve for Acceleration

Rearrange and solve for \( a \): \[ a = -\frac{(9.9)^2}{2 \times 32.0} \] Calculate \( a \): \[ a = -\frac{98.01}{64} = -1.53125 \, \mathrm{m/s^2} \]. The negative sign indicates deceleration.

Step 4: Relate Friction and Acceleration

Knowing that the only force acting horizontally is the friction force, it equals the deceleration force. Use \( f = \mu mg = ma \) to relate friction to acceleration, \( \mu g = |a| \).

Step 5: Solve for the Coefficient of Friction

Substitute \( g = 9.81 \, \mathrm{m/s^2} \) and \( a \): \( \mu = \frac{|a|}{g} = \frac{1.53125}{9.81} \). Calculate \( \mu \): \[ \mu = 0.156 \].

Key Concepts

Kinematic Equations

Kinematic equations are indispensable when it comes to solving physics problems involving motion. They allow us to calculate unknown variables when an object moves under uniform acceleration. If any three of the five variables are known (initial velocity, final velocity, acceleration, time, and displacement), we can determine the other two.
A standard kinematic equation that relates these variables is:

• \( v_f^2 = v_i^2 + 2ad \)

Here, \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the displacement. In our hockey puck problem, since the puck comes to rest, \( v_f \) is zero. Using the kinematic equation, we substitute known values to find the deceleration. Understanding how to utilize kinematic equations simplifies the process of analyzing uniform motion. Be sure to always rearrange the equation to isolate the desired variable.

Deceleration

Deceleration is the rate at which an object slows down. It's essentially negative acceleration. In our exercise, we dealt with a hockey puck that decelerates as it moves across the ice. This happens because of the frictional force acting opposite to the direction of its motion.
To find the deceleration, we rearrange the kinematic equation mentioned earlier:

• \( a = -\frac{(v_i)^2}{2d} \)

After substituting our known values, we calculate the deceleration to be \(-1.53125 \ \mathrm{m/s}^2\). The negative sign simply indicates that the puck is slowing down, not speeding up. Deceleration is crucial for understanding how forces like friction impact an object’s motion over time. It offers insight into how quickly an object comes to a halt.

Physics Problem Solving

Physics problem solving is a structured approach where steps are systematically followed to arrive at a solution. It's important to start by identifying known values and exactly what is being asked. This includes understanding key terms like initial velocity, force, and friction.
Once you've organized what you know, select the right formula or equation to aid with your solution. Substitute the values you know into the equation. Don't forget to watch out for units, as they can impact your final answer.

• Check units for consistency
• Understand the significance of each calculation step
• Troubleshoot mistakes by verifying each part of the equation

In solving the hockey puck problem, applying these systematic steps helped us conclude with the coefficient of friction, \( \mu = 0.156 \). This problem-solving approach not only builds confidence in handling similar exercises but also strengthens overall analytical skills.