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Chapter 4: Problem 46

A hockey puck struck by a hockey stick is given an initial speed \(v_{0}\) in the positive \(x\) -direction. The coefficient of kinetic friction between the ice and the puck is \(\mu_{k^{\prime}}\) (a) Obtain an expression for the acceleration of the puck. (b) Use the result of part (a) to obtain an expression for the distance \(d\) the puck slides. The answer should be in terms of the variables \(v_{0}, \mu\), and \(g\) only.

Short Answer

Expert verified
The acceleration of the puck is \(a = -\mu_{k'} g\) and the distance the puck slides before it stops is \(d = -v_0^2 / 2\mu_{k'}g\).

Step by step solution

01

Determine Acceleration

The only force exerted on the hockey puck comes from friction, which is given by the formula \(F_{f}= \mu_{k'} m g\), where \(m\) is the mass of the puck, \(g\) is the acceleration due to gravity, and \(\mu_{k'}\) is the coefficient of kinetic friction. Since force is mass times acceleration \(F_{f}=ma\), we can say \(\mu_{k'} m g=ma\). Solving for acceleration, \(a = \mu_{k'} g\) and since the direction is opposite to motion, \(a = - \mu_{k'} g\).
02

Use kinematics to express distance

We use the kinematic equation \(v_f^2 = v_0^2 + 2a d\) where:- \(v_f = 0\) (the final speed is zero because the puck stops)- \(v_0\) is the initial speed - \(a\) is the acceleration - \(d\) is the distance By substituting values in, \(0 = v_0^2 + 2(-\mu_{k'}g)d\), we get \(\mu_{k'}gd = -v_0^2 / 2\). Solving for \(d\), we obtain \(d = -v_0^2 / 2\mu_{k'}g\).

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Key Concepts

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

Coefficient of Kinetic Friction
Understanding the coefficient of kinetic friction is crucial when studying the motion of objects in physics. It is a dimensionless scalar that represents the amount of friction existing between two moving surfaces. In our hockey puck example, it is represented by \(\mu_k'\). When an object like the puck is sliding on ice, the coefficient of kinetic friction will determine how fast it slows down.

The formula to calculate the kinetic friction force involves this coefficient: \(F_f = \mu_k' m g\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(\mu_k'\) is the coefficient of kinetic friction. A higher \(\mu_k'\) means a rougher surface and a quicker deceleration of the moving object. Conversely, a low \(\mu_k'\) would mean a smoother surface and slower deceleration.
Newton's Second Law
Newton's Second Law is fundamental to understanding motion in physics. It states that the force acting upon an object is equal to the mass of that object multiplied by its acceleration \(F=ma\). This law is pivotal when it comes to calculating the acceleration of an object when forces are applied.

In our scenario, the only horizontal force acting on the hockey puck is the force of kinetic friction. The force of friction is ultimately what decelerates the puck, allowing us to apply Newton's Second Law to establish the relationship \(F_f = \mu_k' m g = ma\), hence \(a = \mu_k' g\). Notice that the force of friction is negative since it acts opposite to the direction of motion, aligning with the directional nature of Newton's Second Law.
Kinematic Equations
The kinematic equations describe the motion characteristics of a moving body under the action of constant acceleration. They are essential tools for solving problems in classical mechanics involving motion. One such equation is \(v_f^2 = v_0^2 + 2ad\), where \(v_f\) represents final velocity, \(v_0\) is initial velocity, \(a\) is constant acceleration, and \(d\) is the distance covered.

By using this equation, we can determine the distance the hockey puck travels before coming to a stop. After setting \(v_f=0\) since the puck stops, and solving for \(d\), the formula becomes \(d = -v_0^2 / 2\mu_k'g\). This kinematic equation, combined with the coefficient of kinetic friction and the understanding of acceleration, unveils the direct relationship between initial velocity, friction, and the distance an object will travel.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \(g\), is the acceleration of an object due to Earth's gravitational pull. In physics, it is a fundamental concept that affects every object with mass. The standard value for Earth's gravity is \(9.81 m/s^2\), although this can slightly vary depending on where you are on the globe.

In calculations involving kinetic friction and moving objects on the Earth's surface, \(g\) is used to determine the normal force which in turn helps in calculating the frictional forces. In the case of the hockey puck, the acceleration due to gravity is a key part of the calculations for both the frictional force \(F_f\) and the acceleration \(a\) of the puck, as it directly affects the normal force exerted by the ice on the puck, thus influencing how fast the puck slows down on the ice.

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Most popular questions from this chapter

A magician pulls a tablecloth from under a \(200-\mathrm{g} \mathrm{mug}\) located \(30.0 \mathrm{~cm}\) from the edge of the cloth. The cloth exerts a friction force of \(0.100 \mathrm{~N}\) on the mug and is pulled with a constant acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than \(30 \mathrm{~cm}\) relative to the tabletop during the process.

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the \(\mathrm{cab}\), if the truck stops suddenly in an accident or even in braking. Assume, for example, a \(10000-\mathrm{kg}\) load sits on the flatbed of a 20 \(000-\mathrm{kg}\) truck moving at \(12.0 \mathrm{~m} / \mathrm{s}\). Assume the load is not tied down to the truck and has a coefficient of static friction of \(0.500\) with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution?

A frictionless plane is \(10.0 \mathrm{~m}\) long and inclined at \(35.0^{\circ}, \mathrm{A}\) sled starts at the bottom with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed \(v_{e}\). Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled.

(a) An clevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?^{2}\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w ?\) (c) When the elevator is moving upward, but the acceleration is downward, which is greater, Tor w? (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\), Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find \(T\). Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\), Find \(T\). Is your answer consistent with your answer to part (c)?

A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of \(1.2 \mathrm{~m} / \mathrm{s}\) in \(0.80 \mathrm{~s} .\) The elevator travels with this constant speed for \(5.0 \mathrm{~s}\), undergoes a uniform negative acceleration for \(1.5 \mathrm{~s}\), and then comes to rest. What does the spring scale register (a) before the elevator starts to move? (b) During the first \(0.80 \mathrm{~s}\) of the elevator's ascent? (c) While the elevator is traveling at constant speed? (d) During the elevator's negative acceleration?
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