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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 \(\mu_{k^{*}}\) (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 \(v_{0}, \mu_{k}\), and \(g\) only.

Short Answer

Expert verified
The acceleration of the puck is \(a = -\mu_{k^{*}}g\) and the distance the puck slides is \(d=\frac{v_{0}^{2}}{\mu_{k^{*}}g}\).

Step by step solution

01

Understanding friction and acceleration

It is important to note that when the hockey puck slides on the ice, the only horizontal force acting on it is the frictional force, which acts in the opposite direction to the puck's motion due to kinetic friction.Using the Newton's second law of motion \(F_{net}=ma\), where \(F_{net}\) is the net force, \(m\) is the mass of the hockey puck and \(a\) is its acceleration, we have \(F_{friction}=ma\). Considering that the frictional force can also be expressed as \(F_{friction}=\mu_{k}F_{n}\), where \(F_{n}\) is the normal force which in this case equals \(mg\) as the puck travels horizontally and \(\mu_{k^{*}}\) is the coefficient of kinetic friction.
02

Formulate an expression for acceleration

Setting the two expressions for \(F_{friction}\) equal to each other we get: \(\mu_{k^{*}}F_{n}=ma\), replace \(F_{n}\) by \(mg\), to get: \(\mu_{k^{*}}mg=ma\). Solving this equation for \(a\), we find the expression for the puck's acceleration: \(a=-\mu_{k^{*}}g\). Here the negative sign indicates that the acceleration (deceleration indeed) is in the direction opposite to the initial velocity.
03

Derive an expression for the distance travelled

The deceleration of puck can be defined as the change in velocity over the change in time \(-v_{0}/t\), according to the definition of acceleration. By replacing \(a\) in \(a=-\mu_{k^{*}}g\) we get \(-v_{0}/t=-\mu_{k^{*}}g\). Thus, the time \(t\) the puck takes to stop is \(t=v_{0}/ (\mu_{k^{*}}g)\). We can replace this value of \(t\) in the equation for distance travelled with constant acceleration \(d=v_{0}t+0.5at^{2}\). Considering \(a=-\mu_{k^{*}}g\) and that the puck stops, the distance travelled will be \(d=v_{0}t\). We then obtain that \(d=\frac{v_{0}^{2}}{\mu_{k^{*}}g}\).

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Key Concepts

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

Kinetic Friction
When a hockey puck slides across the ice, the primary force that opposes its motion is the kinetic friction. This type of friction occurs between surfaces in relative motion. In our scenario, the puck and the ice surface are the contact surfaces. Kinetic friction's role is pivotal because it determines how quickly the puck stops. It acts opposite to the direction of motion, causing the puck to decelerate. The force of kinetic friction can be calculated using the formula:
  • \(F_{friction} = \mu_{k} \times F_{n}\)
where \(\mu_{k}\) is the coefficient of kinetic friction and \(F_{n}\) is the normal force. On a flat surface like ice, the normal force is equal to the weight of the puck \(mg\). Therefore, the frictional force is:
  • \(F_{friction} = \mu_{k} \times mg\)
Understanding kinetic friction helps us gauge how different surfaces affect the motion of objects and, specifically, how an ice surface allows a puck to slide with minimal resistance.
Acceleration
Acceleration is the rate at which an object's velocity changes. For the hockey puck, the acceleration is essential to understand because it tells us how quickly it slows down due to kinetic friction. According to Newton's Second Law, acceleration can be determined when we know the net force acting on an object and its mass:
  • \(F_{net} = ma\)
In the case of our puck sliding on ice, the only horizontal force acting is the frictional force. Thus, acceleration \(a\) can be calculated from:
  • \(F_{friction} = ma\)
Substituting the frictional force expression \(\mu_{k} mg = ma\), we solve for acceleration:
  • \(a = -\mu_{k}g\)
The negative sign indicates that the acceleration is actually a deceleration, as it opposes the initial velocity. This shows how understanding friction allows us to predict how quickly an object will stop.
Distance Travelled
The distance a moving object travels before coming to a stop is affected by its initial velocity and the forces acting against it, like friction. For the hockey puck, calculating the distance travelled involves determining the time it takes to stop and how far it goes in that time frame.To find the stopping time \(t\), we use the relationship:
  • \(t = \frac{v_{0}}{\mu_{k}g}\)
where \(v_{0}\) is the initial speed, \(\mu_{k}\) is the friction coefficient, and \(g\) is the acceleration due to gravity.Next, using the distance formula for an object under constant acceleration, we find:
  • \(d = v_{0}t + 0.5at^{2}\)
Since the puck stops, the formula simplifies to:
  • \(d = \frac{v_{0}^{2}}{\mu_{k}g}\)
This expression shows how initial speed and surface conditions (embodied in \(\mu_{k}\)) dictate the slide distance. It illustrates the practical influence of kinetic friction on motion.

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Most popular questions from this chapter

A frictionless plane is \(10.0 \mathrm{~m}\) long and inclined at \(35.0^{\circ} .\) 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 \(v_{i}\). 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.

An inquisitive physics student, wishing to combine pleasure with scientific inquiry, rides on a roller coaster sitting on a bathroom scale. (Do not try this yourself on a roller coaster that forbids loose, heavy packages.) The bottom of the seat in the roller-coaster car is in a plane parallel to the track. The seat has a perpendicular back and a seat belt that fits around the student's chest in a plane parallel to the bottom of the seat. The student lifts his feet from the floor so that the scale reads his weight, \(200 \mathrm{lb}\), when the car is horizontal. At one point during the ride, the car zooms with negligible friction down a straight slope inclined at \(30.0^{\circ}\) below the horizontal. What does the scale read at that point?

A crate of mass \(45.0 \mathrm{~kg}\) is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is \(0.350\), and the coefficient of kinetic friction is \(0.320\). (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed? (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?

A crate of weight \(F_{g}\) is pushed by a force \(\overrightarrow{\mathbf{P}}\) on a horizontal floor as shown in Figure \(\mathrm{P} 4.83\). The coefficient of static friction is \(\mu_{s}\), and \(\overrightarrow{\mathbf{P}}\) is directed at angle \(\theta\) below the horizontal. (a) Show that the minimum value of \(P\) that will move the crate is given by $$ P=\frac{\mu_{s} F_{g} \sec \theta}{1-\mu_{s} \tan \theta} $$ (b) Find the condition on \(\theta\) in terms of \(\mu_{s}\) for which motion of the crate is impossible for any value of \(P\).

The parachute on a race car of weight \(8820 \mathrm{~N}\) opens at the end of a quarter-mile run when the car is traveling at \(35 \mathrm{~m} / \mathrm{s}\). What total retarding force must be supplied by the parachute to stop the car in a distance of \(1000 \mathrm{~m}\) ?
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