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A hockey player hits a puck with his stick, giving the puck an initial speed of \(5.0 \mathrm{~m} / \mathrm{s}\). If the puck slows uniformly and comes to rest in a distance of \(20 \mathrm{~m}\), what is the coefficient of kinetic friction between the ice and the puck?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.064.

Step by step solution

01

Understand the Known Values

We know the initial velocity of the puck \(v_i = 5.0 \mathrm{~m/s}\) and the final velocity \(v_f = 0 \mathrm{~m/s}\) since the puck comes to rest. The distance the puck travels before stopping is \(d = 20 \mathrm{~m}\). We need to find the coefficient of kinetic friction \(\mu_k\).
02

Use the Equation of Motion to Find Acceleration

The equation of motion that relates initial velocity, final velocity, acceleration, and distance is given by:\[ v_f^2 = v_i^2 + 2a d \]Plugging in the known values, we solve for \(a\):\[ 0 = (5.0)^2 + 2a(20) \]\[ 0 = 25 + 40a \]\[ 40a = -25 \]\[ a = -\frac{25}{40} = -0.625 \mathrm{~m/s^2} \]
03

Relate Acceleration to Friction

Using Newton's second law, the frictional force \(F_f\) is given by \(F_f = ma\), where \(m\) is the mass of the puck. However, since we are looking for the coefficient of kinetic friction, we can use \(F_f = \mu_k N\), where \(N\) is the normal force, \(N = mg\). Thus:\[ \mu_k mg = ma \]\[ \mu_k g = |a| \]Since the motion is horizontal, and assuming a frictionless surface, we divide acceleration by gravity \(g = 9.8 \mathrm{~m/s^2}\) to find \(\mu_k\):\[ \mu_k = \frac{|a|}{g} = \frac{0.625}{9.8} \]
04

Calculate the Coefficient of Kinetic Friction

Substitute the known values to find \(\mu_k\):\[ \mu_k = \frac{0.625}{9.8} = 0.064 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of Motion
The equation of motion is a mathematical tool that helps us link various elements of motion, such as velocity, acceleration, and distance. This particular equation to find the acceleration when the puck is in motion is written as:
  • \[ v_f^2 = v_i^2 + 2a d \]
By using this equation, we can understand how an object moves over time. It shows the relationship between the initial velocity \(v_i\), the final velocity \(v_f\), the acceleration \(a\), and the distance covered \(d\). By substitution, we can solve for the unknown parameter, such as acceleration in this case. If you rearrange the equation to solve for \(a\), you get:
  • \[ a = \frac{v_f^2 - v_i^2}{2d} \]
Plugging in the known values is straightforward. This helps to determine how quickly an object's velocity changes with respect to time when some forces act upon it.
Newton's Second Law
Newton's second law is a fundamental principle of physics that relates to how forces affect the motion of objects. It can be expressed through the equation:
  • \[ F = ma \]
This law tells us that force is the product of an object's mass \(m\) and its acceleration \(a\). In the context of our problem, this principle is used to relate the frictional force experienced by the puck to its mass and acceleration. Newton's second law explains how much force is necessary to change an object's velocity, helping us calculate the force required to make the puck stop.
When applied to the hockey puck, the force due to friction that opposes its motion can also be described in terms of kinetic friction as:
  • \[ F_f = \mu_k N \]
where \( \mu_k \) is the coefficient of kinetic friction and \(N\) is the normal force, provided by the puck's weight.
Acceleration
Acceleration is a measure of how fast an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. The puck in this problem decelerates uniformly, reducing its speed to zero. The calculation of acceleration involves understanding how quickly the puck comes to rest as it slides over the ice.
  • From the equation of motion, we determine that:\[ a = -0.625 \text{ m/s}^2 \]
The negative sign indicates that the puck is slowing down. Understanding acceleration helps us to analyze whether an object is speeding up, slowing down, or changing direction during its motion. In practical terms, knowing the acceleration is crucial in calculating other dynamics like the distance required to stop an object, or the force exerted during its motion.
Coefficient of Friction
The coefficient of friction is a unitless value that represents the ratio of the force of friction between two bodies and the force pressing them together. In our scenario, it tells us how much frictional resistance the puck experiences as it glides across the ice. The coefficient of kinetic friction \( \mu_k \) can be calculated using the formula:
  • \[ \mu_k = \frac{|a|}{g} \]
where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\). In the case of the puck, using the calculated acceleration, we find:
  • \[ \mu_k = \frac{0.625}{9.8} \approx 0.064 \]
This tells us how grippy or slippery the ice is in relation to the puck. A lower value of \(\mu_k\) indicates less friction, hence a smoother sliding surface. Understanding the coefficient of friction aids in designing surfaces and materials for specific needs, whether for maximizing speed or for providing stability and control.

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Most popular questions from this chapter

The coefficients of static and kinetic friction between a \(50.0-\mathrm{kg}\) box and a horizontal surface are 0.500 and 0.400 respectively. (a) What is the acceleration of the object if a 250-N horizontal force is applied to the box? (b) What is the acceleration if the applied force is \(235 \mathrm{~N}\) ?

(a) When an object is on an inclined plane, the normal force exerted by the inclined plane on the object is (1) less than, (2) equal to, (3) more than the weight of the object. Why? (b) For a \(10-\mathrm{kg}\) object on a \(30^{\circ}\) inclined plane, what are the object's weight and the normal force exerted on the object by the inclined place?

When a horizontal force of \(300 \mathrm{~N}\) is applied to a 75.0 \(\mathrm{kg}\) box, the box slides on a level floor, opposed by a force of kinetic friction of \(120 \mathrm{~N}\). What is the magnitude of the acceleration of the box?

A \(2.50-\mathrm{kg}\) block is placed on a rough surface inclined at \(30^{\circ} .\) The block is propelled and launched at a speed of \(1.60 \mathrm{~m} / \mathrm{s}\) down the incline and comes to rest after sliding \(1.10 \mathrm{~m} .\) (a) Draw the free-body diagram of the block while it is sliding. Also indicate your coordinate system axes. (b) Starting with Newton's second law applied along both axes of your coordinate system, use your free- body diagram to generate two equations. (c) Solve these equations for the coefficient of kinetic friction between the block and the incline surface. [Hint: You will need to first determine the block's acceleration.]

A \(3000-\mathrm{kg}\) truck tows a \(1500-\mathrm{kg}\) car by a chain. If the net forward force on the truck by the ground is \(3200 \mathrm{~N}\), (a) what is the acceleration of the car, and (b) what is the tension in the connecting chain?

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