/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Force Times Time At the local ho... [FREE SOLUTION] | 91Ó°ÊÓ

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Force Times Time At the local hockey rink, a puck with a mass of \(0.12 \mathrm{kg}\) is given an initial speed of \(v_{0}=6.7 \mathrm{m} / \mathrm{s}\). (a) If the coefficient of kinetic friction between the ice and the puck is 0.13 , how much time \(t\) does it take for the puck to come to rest? (b) If the mass of the puck is doubled, does the frictional force \(F\) exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping time of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that \(F t=m v_{0}\). (The significance of this result will be discussed in Chapter \(9,\) where we will see that \(m v\) is the momentum of an object.)

Short Answer

Expert verified
(a) 5.26 s; (b) Increases; (c) Stays the same; (d) Yes, it shows impulse equals momentum change.

Step by step solution

01

Understanding Frictional Force

The frictional force can be calculated using the equation: \( F_{friction} = \mu_k \cdot m \cdot g \), where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass of the puck, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Substitute the given values to find the frictional force.
02

Calculating Frictional Force

Substitute \( \mu_k = 0.13 \), \( m = 0.12 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \) into the friction force equation: \( F_{friction} = 0.13 \times 0.12 \times 9.8 \). This calculates to \( F_{friction} = 0.15288 \, \text{N} \).
03

Find Time Using Newton's Second Law

Use Newton's second law, \( F = m \cdot a \), where \( a \) is the acceleration. Since \( F_{friction} = m \cdot a \), we can solve for \( a = \frac{F_{friction}}{m} \). Substitute \( F_{friction} = 0.15288 \, \text{N} \) and \( m = 0.12 \, \text{kg} \): \( a = \frac{0.15288}{0.12} = 1.274 \, \text{m/s}^2 \).
04

Calculate Time to Stop the Puck

Use the kinematic equation \( v = v_0 + at \) and set \( v = 0 \) to find \( t \), where \( v_0 = 6.7 \, \text{m/s} \) and \( a = -1.274 \, \text{m/s}^2 \). Solving \( 0 = 6.7 - 1.274t \) gives \( t = \frac{6.7}{1.274} = 5.26 \, \text{s} \).
05

Effect of Doubling Mass on Frictional Force (b)

Frictional force depends on the normal force, which is the weight of the puck: \( F_{friction} = \mu_k m g \). Doubling the mass doubles the weight, so the frictional force also doubles.
06

Effect of Doubling Mass on Stopping Time (c)

The stopping force (friction) increases linearly with mass, but so does inertia (resistance to change in motion). Since both the force and resistance double, time remains unchanged when mass is doubled due to cancelling effects.
07

Proving Ft = mv_0 (d)

The change in momentum \( \Delta p = m(v - v_0) = -mv_0 \) as the puck stops. The impulse is equal to the change in momentum: \( F \cdot t = m v_0 \). Because \( a = \frac{F}{m} \), rearranging and substituting gives \( F \cdot t = m (-a) \cdot \frac{v_0}{a} = mv_0 \). This verifies the given relationship.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics that connects the force applied to a body, its mass, and the resulting acceleration. The law is expressed by the formula: \( F = m \cdot a \) where \( F \) is force in newtons, \( m \) is mass in kilograms, and \( a \) is acceleration in meters per second squared. This law implies that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. For the hockey puck in our exercise, the frictional force is the external force causing it to decelerate. By applying the formula, we find the puck's acceleration by dividing the friction force by the puck's mass. This understanding then helps us calculate how long it will take for the puck to stop.
Impulse and Momentum
Impulse and momentum are key concepts in the study of motion, describing how forces affect the movement of objects over time. Momentum, \( p \), is the product of an object's mass and velocity: \[ p = m \cdot v \]It measures how much motion an object has and is conserved in the absence of external forces. Impulse, on the other hand, is the change in an object's momentum, given by the formula:\( F \cdot t = \Delta p \)In the exercise, the impulse provided by friction changes the puck's momentum from \( m \cdot v_0 \) to zero as it comes to rest. The relationship \( F \cdot t = m \cdot v_0 \) indicates that the product of force and time equals the initial momentum, demonstrating how impulse allows us to calculate the stopping time.
Kinematics
Kinematics focuses on the motion of objects without considering the forces that cause the motion. It helps us describe the motion using quantities such as velocity, acceleration, and displacement.In our exercise, we use the kinematic equation:\[ v = v_0 + at \]This equation allows us to solve for the time \( t \) it takes for the puck to come to a stop. By setting the final velocity \( v \) to zero and substituting the initial velocity \( v_0 \) and the calculated acceleration \( a \), we can find the stopping time. This approach ties together the motion characteristics of the puck through the frictional force acting on it, without needing to delve directly into the force concepts.
Coefficient of Friction
The coefficient of friction, \( \mu_k \), is a dimensionless constant that quantifies the frictional resistance between two surfaces in relative motion. It plays a crucial role in determining the frictional force.Given by:\[ F_{friction} = \mu_k \cdot m \cdot g \]this equation shows that the frictional force is directly proportional to both the coefficient of friction and the normal force. In the context of the exercise, the ice rink's coefficient of kinetic friction affects how quickly the puck decelerates. A higher coefficient means more friction and a shorter travel time before stopping, while a lower coefficient would mean less resistance and a longer time to stop. Understanding this parameter helps in predicting and controlling the motion of objects on different surfaces.

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Most popular questions from this chapter

A \(50.0-\mathrm{kg}\) person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of \(15.0^{\circ}\) above the horizontal. Find the tension in the ropes.

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A certain spring has a force constant \(k\). (a) If this spring is cut in half, does the resulting half spring have a force constant that is greater than, less than, or equal to \(k ?\) (b) If two of the original full-length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to \(k\) ?

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