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A horizontal force equal to the object's weight is applied to an object resting on a table. What is the acceleration of the moving object when the coefficient of kinetic friction between the object and floor is 1 (assuming the object is moving in the direction of the applied force). a) zero c) Not enough information is b) \(1 \mathrm{~m} / \mathrm{s}^{2}\) given to find the acceleration.

Short Answer

Expert verified
Answer: The acceleration of the object is 0 m/sĀ².

Step by step solution

01

Determine the Applied Force

To determine the acceleration of the object, we first need to find out the applied force. The weight of the object can be calculated using the formula \(W=mg\), where \(W\) is the weight, \(m\) is the mass, and \(g\) is the gravitational acceleration (\(9.81 \mathrm{~m} / \mathrm{s}^{2}\)). The problem states that the applied force is equal to the object's weight, so we have: \(F_{applied} = W = mg\)
02

Calculate the Friction Force

Now, we need to calculate the friction force experienced by the object when it is moving. The friction force (\(F_{friction}\)) can be calculated using the formula: \(F_{friction} = \mu F_N\), where \(\mu\) is the coefficient of friction and \(F_N\) is the normal force. In this case, since the object is resting on a horizontal surface, the normal force is equal to the object's weight. Therefore, we have: \(F_{friction} = \mu mg\) Given that the coefficient of kinetic friction (\(\mu\)) between the object and the floor is 1, the friction force becomes: \(F_{friction} = 1 \cdot mg = mg\)
03

Apply Newton's Second Law

Now, we can apply Newton's second law to find the acceleration of the moving object. According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration: \(F_{net} = ma\). The net force acting on the object is the difference between the applied force and the friction force. Since the applied force and friction force are equal, we have: \(F_{net} = F_{applied} - F_{friction} = mg - mg = 0\) Using Newton's second law, we can now calculate the acceleration: \(0 = ma\)
04

Determine the Acceleration

Finally, to determine the acceleration of the moving object, we need to solve the equation obtained in the last step: \(0 = ma\) Dividing both sides by \(m\), we get the acceleration: \(a = 0 \mathrm{~m} / \mathrm{s}^{2}\) Therefore, the correct answer is (a) zero.

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Key Concepts

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

Kinetic Friction
Kinetic friction occurs when an object slides or moves over a surface. It acts in the opposite direction to the movement and tries to slow it down.
The frictional force is determined by two main factors:
  • The coefficient of kinetic friction, represented as \( \mu \), which is a dimensionless value specific to the contacting surfaces.
  • The normal force \( F_N \), which is generally the weight of the object when on a horizontal surface.
To calculate the kinetic frictional force, use the formula:\[F_{friction} = \mu F_N \]In our exercise, the coefficient of kinetic friction \( \mu \) is 1. The normal force \( F_N \) equals the weight of the object, \( mg \). Hence, the frictional force becomes:\[F_{friction} = mg \]This knowledge allows us to predict how objects will behave when moving across surfaces, which is crucial for understanding the dynamics and energy requirements in various physical systems.
Net Force
Net force is the overall force acting on an object when all individual forces are combined. It determines the object's acceleration in accordance with Newton's Second Law. In simple terms, if you know the forces acting upon an object, you can determine the net force and, subsequently, predict how the object will move.
For our scenario, we have:
  • An applied force equivalent to the object's weight \( F_{applied} = mg \)
  • The kinetic friction force \( F_{friction} = mg \)
The net force \( F_{net} \) then becomes:\[F_{net} = F_{applied} - F_{friction} = mg - mg = 0 \]Since the forces balance out, the net force is zero. This indicates that the object doesnā€™t speed up or slow down, reaffirming the principle that net force equals mass times acceleration \( F_{net} = ma \). Thus, if \( F_{net} = 0 \), the acceleration must be zero.
Acceleration Calculation
Acceleration measures the rate of change of velocity of an object and is a cornerstone in understanding motion. To find acceleration, we reassure Newton's Second Law:\[F_{net} = ma \]In our problem, after calculating both the applied and frictional forces, we found that they cancel each other out leading to a net force of zero. Plugging this into Newton's equation, we have:\[0 = ma \]To isolate acceleration \( a \), divide both sides by the object's mass \( m \):\[a = \frac{0}{m} = 0 \]As a result, the acceleration is zero. This shows that if no net force acts upon an object, its velocity remains constant. If it was at rest, it stays at rest, and if it was moving, it keeps moving at the same speed. This is a fundamental insight into how forces affect motion.

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Most popular questions from this chapter

Two blocks of equal mass are connected by a massless horizontal rope and resting on a frictionless table. When one of the blocks is pulled away by a horizontal external force \(\vec{F}\) what is the ratio of the net forces acting on the blocks? a) 1: 1 c) 1: 2 b) 1: 1.41 d) none of the above

The density (mass per unit volume) of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\) and the density of seawater is \(1024 \mathrm{~kg} / \mathrm{m}^{3}\). Only \(10.4 \%\) of the volume of an iceberg is above the water's surface. If the volume of a particular iceberg that is above water is \(4205.3 \mathrm{~m}^{3},\) what is the magnitude of the force that the seawater exerts on this iceberg?

A heavy wooden ball is hanging from the ceiling by a piece of string that is attached from the ceiling to the top of the ball. A similar piece of string is attached to the bottom of the ball. If the loose end of the lower string is pulled down sharply, which is the string that is most likely to break?

An engine block of mass \(M\) is on the flatbed of a pickup truck that is traveling in a straight line down a level road with an initial speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The coefficient of static friction between the block and the bed is \(\mu_{s}=0.540 .\) Find the minimum distance in which the truck can come to a stop without the engine block sliding toward the cab.

A chair of mass \(M\) rests on a level floor, with a coef. ficient of static friction \(\mu_{s}=0.560\) between the chair and the floor. A person wishes to push the chair across the floor. He pushes on the chair with a force \(F\) at an angle \(\theta\) below the horizontal. What is the maximum value of \(\theta\) for which the chair will not start to move across the floor?

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