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Show that the acceleration of any object down an incline where friction behaves simply (that is, where \(\left.f_{\mathrm{k}}=\mu_{\mathrm{k}} N\right)\) is \(a=g\left(\sin \theta-\mu_{\mathrm{k}} \cos \theta\right) .\) Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small \(\left(\mu_{\mathrm{k}}=0\right)\)

Short Answer

Expert verified
To show that the acceleration of an object down an incline is \(a = g(\sin \theta - \mu_{\mathrm{k}} \cos \theta)\), we first identify the forces acting on the object (gravitational and frictional forces), and resolve the gravitational force into components parallel and perpendicular to the incline. The normal force is equal to the perpendicular component, and frictional force is calculated using the equation \(F_{\mathrm{f}} = \mu_{\mathrm{k}} N\). The net force is calculated by subtracting the frictional force from the parallel component of the gravitational force. Using Newton's second law, we get the expression for acceleration, which is independent of mass and reduces to \(g\sin \theta\) when friction is negligibly small.

Step by step solution

01

Identify the forces acting on the object on the incline

Consider an object placed on an inclined plane, making an angle \(\theta\) with the horizontal. The object is experiencing a gravitational force perpendicular to the incline, along with the force due to friction between the object and the inclined surface. There are two main forces acting on the object: 1. Gravitational force (\(F_g = mg\)), acting vertically downwards. 2. Frictional force (\(F_{\mathrm{f}} = \mu_{\mathrm{k}} N\)), acting opposite to the direction of motion. Next, we will resolve the gravitational force into components parallel and perpendicular to the incline.
02

Resolve gravitational force

The gravitational force can be resolved into the component parallel (\(F_{g\parallel}\)) and perpendicular (\(F_{g\perp}\)) to the incline to facilitate our calculation. Using trigonometry, we have: 1. Parallel component: \(F_{g\parallel} = mg \sin \theta\) 2. Perpendicular component: \(F_{g\perp} = mg \cos \theta\)
03

Find the normal force

The object is not moving perpendicular to the incline, as it is in contact with the inclined surface. This indicates that the normal force (\(N\)) is equal to the weight component perpendicular to the incline: \(N = F_{g\perp} = mg \cos \theta\)
04

Determine frictional force

The frictional force (\(F_{\mathrm{f}}\)) can be determined using the equation given in the problem statement: \(F_{\mathrm{f}} = \mu_{\mathrm{k}} N = \mu_{\mathrm{k}} mg \cos \theta\)
05

Calculate the net force on the object

The net force (\(F_{\mathrm{net}}\)) acting on the object down the incline can now be calculated. The only two forces acting along the incline are the gravitational force component parallel to the incline \(F_{g\parallel}\) and the frictional force \(F_{\mathrm{f}}\): \(F_{\mathrm{net}} = F_{g\parallel} - F_{\mathrm{f}} = mg\sin \theta - \mu_{\mathrm{k}} mg\cos\theta\)
06

Calculate acceleration of the object down the incline

Using Newton's second law of motion, \(F = ma\), we can calculate the acceleration (\(a\)) of the object: \(ma = F_{\mathrm{net}} = mg\sin \theta - \mu_{\mathrm{k}} mg\cos\theta\) Notice that mass (\(m\)) is a common factor in this expression; hence, we can divide both sides by the object's mass: \(a = g(\sin \theta - \mu_{\mathrm{k}} \cos \theta)\) This expression shows that the acceleration of the object down the incline is independent of mass and is given by: \(a = g(\sin \theta - \mu_{\mathrm{k}} \cos \theta)\)
07

Validate with negligibly small friction

Now, we will check if the expression given in the problem statement for negligibly small friction holds. If friction is negligibly small, then \(\mu_{\mathrm{k}} = 0\): \(a = g(\sin \theta - \mu_{\mathrm{k}} \cos \theta) = g(\sin \theta - 0 \cdot \cos \theta) = g\sin \theta\) As expected, the formula for acceleration reduces to \(g\sin \theta\) when friction becomes negligibly small. Therefore, the given expression for acceleration down an incline is validated.

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Key Concepts

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

Gravitational Force
When an object rests on an inclined plane, gravity plays a key role in its motion. The gravitational force is the force of attraction that the Earth exerts on the object. This force acts vertically downwards, towards the center of the Earth, and can be represented by the equation:- Gravitational Force: \( F_g = mg \)where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately 9.81 m/s² on Earth. To analyze its effects on an incline, we break it into two components:
  • Parallel Component: aligns with the direction of potential motion down the slope, described by \( F_{g\parallel} = mg \sin \theta \)
  • Perpendicular Component: directed into the plane, described by \( F_{g\perp} = mg \cos \theta \)
These components help us understand how the gravitational force affects the object’s movement over the incline.
Frictional Force
Frictional force is the resistance that surfaces exert when they move across each other. On an inclined plane, this force works opposite to the object's motion. The size of the frictional force depends on two key factors: the nature of the surfaces in contact and the normal force exerted by the surface.- Frictional Force Equation: \( F_{\mathrm{f}} = \mu_{\mathrm{k}} N \)Here, \( \mu_{\mathrm{k}} \) is the coefficient of kinetic friction, which is a value that quantifies the friction between the specific surfaces in contact, and \( N \) is the normal force. The frictional force aims to prevent or slow down potential motion down the incline by opposing the gravitational force component that acts along the slope.
Newton's Second Law
Newton’s Second Law of Motion is a crucial principle that relates the net force acting on an object to its mass and acceleration. It is expressed as:- \( F = ma \)Where \( F \) is the net force on the object, \( m \) is the object's mass, and \( a \) is its acceleration. In the context of an inclined plane, the net force is the result of both the parallel component of the gravitational force and the frictional force. By applying Newton's Second Law, we can compute the object's acceleration down the inclined plane, leading to the expression:- \( a = g(\sin \theta - \mu_{\mathrm{k}} \cos \theta) \)This calculation shows that the acceleration is determined by both the slope of the incline and the frictional characteristics, and interestingly, it does not depend on the object's mass.
Normal Force
The normal force is an essential concept when discussing inclined planes. It is the perpendicular contact force exerted by a surface to support the weight of an object resting on it. On an incline, the normal force differs from the gravitational force because the object does not move vertically but remains pressed against the plane.- Normal Force Equation: \( N = mg \cos \theta \)This force counteracts the perpendicular component of the gravitational force, ensuring that the object does not move through the surface. The normal force is significant because it directly influences the frictional force by serving as one of its multiplying factors in the equation for friction: \( F_{\mathrm{f}} = \mu_{\mathrm{k}} N \). Understanding the normal force is crucial for accurately calculating other forces like friction that affects an object's motion on the incline.
Acceleration Down an Incline
The acceleration of an object on an inclined plane is influenced by multiple factors, primarily the gravitational pull, friction, and the angle of the plane. Through the step-by-step computation, we find that the object's acceleration is unaffected by its mass and can be represented with the formula:- Acceleration Equation: \( a = g(\sin \theta - \mu_{\mathrm{k}} \cos \theta) \)Here, \( g \) is the gravitational acceleration, \( \theta \) is the angle of the incline, and \( \mu_{\mathrm{k}} \) is the kinetic friction coefficient. This equation reflects how the plane's steepness and surface characteristics, rather than the object's mass, primarily determine acceleration. In scenarios where friction is minimal or negated, \( \mu_{\mathrm{k}} \) becomes zero, simplifying the formula to \( a = g\sin \theta \), which accounts for gravity's sole influence on the movement.

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Most popular questions from this chapter

In a later chapter, you will find that the weight of a particle varies with altitude such that \(w=\frac{m g r_{0}^{2}}{r^{2}}\) where \(r_{0}\) is the radius of Earth and \(r\) is the distance from Earth's center. If the particle is fired vertically with velocity \(v_{0}\) from Earth's surface, determine its velocity as a function of position \(r\). (Hint: use \(a d r=v d v,\) the rearrangement mentioned in the text.)

A crate of mass \(100.0 \mathrm{kg}\) rests on a rough surface inclined at an angle of \(37.0^{\circ}\) with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is \(80 \%\) of that for the static case. (a) What is the coefficient of static friction? (b) What is the maximum force that can be applied upward along the plane on the rope and not move the block? (c) With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? (d) If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? (e) Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?

A student is attempting to move a 30-kg mini-fridge into her dorm room. During a moment of inattention, the mini-fridge slides down a 35 degree incline at constant speed when she applies a force of \(25 \mathrm{N}\) acting up and parallel to the incline. What is the coefficient of kinetic friction between the fridge and the surface of the incline?

Modern roller coasters have vertical loops like the one shown here. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. (a) What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15.0 \mathrm{m}\) and the downward acceleration of the car is \(1.50 \mathrm{g}\) ? (b) How high above the top of the loop must the roller coaster start from rest, assuming negligible friction? (c) If it actually starts \(5.00 \mathrm{m}\) higher than your answer to (b), how much energy did it lose to friction? Its mass is \(1.50 \times 10^{3} \mathrm{kg}\)

A small space probe is released from a spaceship. The space probe has mass \(20.0 \mathrm{kg}\) and contains \(90.0 \mathrm{kg}\) of fuel. It starts from rest in deep space, from the origin of a coordinate system based on the spaceship, and burns fuel at the rate of \(3.00 \mathrm{kg} / \mathrm{s}\). The engine provides a constant thrust of \(120.0 \mathrm{N}\). (a) Write an expression for the mass of the space probe as a function of time, between 0 and 30 seconds, assuming that the engine ignites fuel beginning at \(t=0 .\) (b) What is the velocity after \(15.0 \mathrm{s}\) ? (c) What is the position of the space probe after \(15.0 \mathrm{s}\), with initial position at the origin? (d) Write an expression for the position as a function of time, for \(t>30.0 \mathrm{s}\)

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