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(II) A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.15 and the push imparts an initial speed of \(3.5 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
The box will slide approximately 4.17 meters.

Step by step solution

01

Identify the Forces

First, identify the forces acting on the box. The major forces are the kinetic friction and the initial push. The force of kinetic friction is what slows the box down as it slides.
02

Use the Formula for Frictional Force

Calculate the force of kinetic friction using the formula: \[ f_k = \mu_k \, N \] where \( \mu_k = 0.15 \) is the coefficient of kinetic friction, and \( N = mg \) is the normal force, with \( m \) being the mass and \( g = 9.8 \, \text{m/s}^2 \) being the acceleration due to gravity.
03

Set Up the Equation of Motion

Apply Newton's second law and consider only the horizontal component, since the box is sliding horizontally. Using \[ F_{net} = ma = -f_k \] yields the following equation: \[ ma = -\mu_k mg \].Cancel \( m \) from both sides of the equation to obtain the acceleration: \[ a = -\mu_k g \]
04

Calculate the Deceleration

Plug the given values into the formula for acceleration obtained:\[ a = -0.15 \times 9.8 = -1.47 \, \text{m/s}^2 \].This is the deceleration due to kinetic friction.
05

Use Kinematic Equation to Find Distance

Now use the kinematic equation: \[ v^2 = u^2 + 2as \] where \( v = 0 \) m/s (final velocity when the box stops),\( u = 3.5 \) m/s (initial speed), and \( a = -1.47 \) m/s². Solve for \( s \):\[ 0 = (3.5)^2 + 2(-1.47)s \]
06

Solve for Distance

Rearrange and solve for \( s \):\[ s = \frac{(3.5)^2}{2 \times 1.47} \approx 4.166 \].Thus, the box slides approximately 4.17 meters before coming to rest.

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Key Concepts

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

Kinematic Equation
The kinematic equation is an essential tool utilized in physics to describe the motion of objects. These equations relate various motion parameters, such as velocity, acceleration, time, and displacement, which can provide crucial insights into the dynamics of moving objects.
The equation relevant to the exercise is given by:
  • \( v^2 = u^2 + 2as \)
Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance traveled. In the context of our exercise, the box starts with an initial speed of 3.5 m/s and eventually comes to a stop. Thus, \( v = 0 \) m/s.

By substituting the known values into this kinematic equation, we can solve for \( s \), which represents the stopping distance of the box. This method is common in physics problems where objects are subjected to constant acceleration or deceleration, as it allows us to understand how far an object will travel before stopping.
Newton's Second Law
Newton's Second Law of Motion provides a fundamental principle for understanding how forces affect the motion of objects. It states that the net force acting on an object is equal to the product of its mass and acceleration:
  • \( F_{net} = ma \)
This law helps us comprehend the box's motion in the problem, where the key force to consider is the kinetic friction that opposes the initial push.
The equation for the net force here includes the frictional force, thus leading to:
  • \( ma = -f_k \)
Substituting \( f_k = \mu_k mg \) into the equation gives us:
  • \( ma = -\mu_k mg \)
Canceling the mass \( m \) from both sides simplifies this to determine the acceleration due to friction, \( a = -\mu_k g \). This simplification is crucial as it shows that the deceleration depends only on the coefficient of friction and gravity, independent of mass.
Coefficient of Friction
The coefficient of friction is a dimensionless scalar that quantifies the frictional force between two surfaces in contact. It is critical for determining the deceleration caused by friction as an object moves across a surface.
Kinetic friction, a type of friction that acts on moving objects, is defined by:
  • \( f_k = \mu_k N \)
where \( \mu_k \) represents the coefficient of kinetic friction, and \( N \) is the normal force. For horizontal surfaces, the normal force \( N \) equals \( mg \), the weight of the object.

In our exercise, \( \mu_k = 0.15 \), which tells us how much frictional resistance the floor offers against the sliding box. Understanding this value is crucial, as it directly affects the box's deceleration and therefore how far it will slide. A higher coefficient would lead to greater friction, resulting in a shorter sliding distance.

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Most popular questions from this chapter

A 2.0 -kg silverware drawer does not slide readily. The owner gradually pulls with more and more force, and when the applied force reaches \(9.0 \mathrm{N},\) the drawer suddenly opens, throwing all the utensils to the floor. What is the coefficient of static friction between the drawer and the cabinet?

(II) \((a)\) Show that the minimum stopping distance for an automobile traveling at speed \(v\) is equal to \(v^{2} / 2 \mu_{\mathrm{s}} g,\) where \(\mu_{\mathrm{s}}\) is the coefficient of static friction between the tires and the road, and \(g\) is the acceleration of gravity. (b) What is this distance for a \(1200-\mathrm{kg}\) car traveling \(95 \mathrm{~km} / \mathrm{h}\) if \(\mu_{\mathrm{s}}=0.65 ?\) (c) What would it be if the car were on the Moon (the acceleration of gravity on the Moon is about \(g / 6\) ) but all else stayed the same?

(II) A 1280 -kg car pulls a 350 -kg trailer. The car exerts a horizontal force of \(3.6 \times 10^{3} \mathrm{~N}\) against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.15 for the trailer.

(II) A coin is placed \(12.0 \mathrm{~cm}\) from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 35.0 rpm (revolutions per minute) is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

(II) Police investigators, examining the scene of an accident involving two cars, measure 72 -m-long skid marks of one of the cars, which nearly came to a stop before colliding. The coefficient of kinetic friction between rubber and the pavement is about \(0.80 .\) Estimate the initial speed of that car assuming a level road.

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