/*! 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 108 Shown below is a 10.0 -kg block ... [FREE SOLUTION] | 91Ó°ÊÓ

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Shown below is a 10.0 -kg block being pushed by a horizontal force \(\overrightarrow{\mathbf{F}}\) of magnitude \(200.0 \mathrm{N}\). The coefficient of kinetic friction between the two surfaces is 0.50. Find the acceleration of the block.

Short Answer

Expert verified
The acceleration of the 10-kg block being pushed by a horizontal force of 200 N with a coefficient of kinetic friction of 0.50 is \(15.1\ m/s^2\).

Step by step solution

01

Identify the forces acting on the block

There are two forces that are acting on the block: the horizontal force (F) and the frictional force (F_friction).
02

Calculate the frictional force

The frictional force can be calculated using the formula: \[F_{friction} = μ * F_{normal}\] Where \(F_{normal}\) is the normal force, and in this case, it is equal to the weight of the block since there's no vertical motion: \[F_{normal} = m * g\] \[F_{normal} = 10 \ kg * 9.8 \ m/s^2\] \[F_{normal} = 98 \ N\] Now, with the given coefficient of kinetic friction (μ), we can find the frictional force: \[F_{friction} = 0.50 * 98 \ N\] \[F_{friction} = 49 \ N\]
03

Calculate the net force acting on the block

To find the net force acting on the block, subtract the frictional force from the applied force: \[F_{net} = F - F_{friction}\] \[F_{net} = 200 \ N - 49 \ N\] \[F_{net} = 151 \ N\]
04

Calculate the acceleration of the block using Newton's second law

Now we can use Newton's second law of motion, which states: \[F_{net} = m * a\] Where F is the net force, m is the mass of the object, and a is the acceleration of the object. Therefore, to find the acceleration, we rearrange the formula: \[a = \frac{F_{net}}{m}\] Plug in the values: \[a = \frac{151 \ N}{10 \ kg}\] \[a = 15.1\ m/s^2\] The acceleration of the 10-kg block is 15.1 m/s².

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Key Concepts

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

Newton's Second Law of Motion
Newton's Second Law of Motion is a fundamental principle that describes the relationship between the net force acting on an object and its acceleration. In essence, this law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This relationship is summed up in the equation:

\[\begin{equation}F_{net} = m \times a\tag{1}\end{equation}\]In the context of our exercise, after calculating the net force acting on the block by considering both the applied force and the opposing force of friction, we use equation (1) to determine the block's acceleration. The greater the net force, the greater the acceleration will be, and vice versa. This concept is crucial for understanding how different forces influence the motion of objects.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, typically denoted as \(\mu_k\), is a dimensionless scalar value that represents the ratio of the force of kinetic friction between two bodies in relative motion to the normal force pressing them together. It is a measure of how much frictional resistance is encountered when one surface slides over another.

Notably, this coefficient is independent of the mass or the area of contact between the two surfaces. It primarily depends on the materials and texture of the surfaces involved. In our exercise, the coefficient of kinetic friction was given as 0.50, which was then utilized in calculation of the force due to friction:\[\begin{equation} F_{friction} = \mu_k \times F_{normal} \tag{2}\end{equation}\]This equation is critical as it provides the means to quantify the frictional force needed to determine the net force on the moving block.
Net Force Calculation
Net force calculation is the process of determining the total force acting on an object, which is the vector sum of all individual forces. This is the force responsible for causing changes in the object’s velocity, which according to Newton’s Second Law, is also the product of the object's mass and acceleration.

In practical situations, such as the block being pushed, we subtract the frictional force, which acts in opposition to the direction of motion, from the applied force to achieve the net force:\[\begin{equation} F_{net} = F_{applied} - F_{friction} \tag{3}\end{equation}\]Hence, the ability to calculate the net force accurately is essential not only for predicting the motion of the object but also for understanding the dynamics of the system as a whole.
Normal Force
The normal force is the perpendicular force exerted by a surface against an object resting on it. It's the counteracting force to the object's weight when the object is not accelerating vertically. In most basic scenarios on flat surfaces, the magnitude of the normal force (\(F_{normal}\)) is equal to the weight of the object (\(mg\)), which results from the acceleration due to gravity (\(g\)) pulling the object downwards.

The normal force is a reactionary force, meaning it adjusts based on the applied forces to ensure that there’s no vertical acceleration of the object. For the problem, the normal force calculation is straightforward:

\[\begin{equation}F_{normal} = m \times g \tag{4}\end{equation}\]This determines the intensity of the frictional force since the normal force is one of the elements needed to calculate friction. It’s important to note that normal force can change with the angle of the surface but remains constant for horizontal surfaces, as demonstrated in the exercise.

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Most popular questions from this chapter

A student's backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at \(3.8 \mathrm{m} / \mathrm{s}^{2},\) the scale reads 60 N. (a) What is the mass of the backpack? (b) What does the scale read if the elevator moves upward while speeding up at a rate \(3.8 \mathrm{m} / \mathrm{s}^{2}\) ? (c) What does the scale read if the elevator moves upward at constant velocity? (d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read?

A car rounds an unbanked curve of radius \(65 \mathrm{m}\). If the coefficient of static friction between the road and car is 0.70, what is the maximum speed at which the car can traverse the curve without slipping?

Show that the acceleration of any object down a frictionless incline that makes an angle \(\theta\) with the horizontal is \(a=g \sin \theta\). (Note that this acceleration is independent of mass.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

. A car of mass \(1000.0 \mathrm{kg}\) is traveling along a level road at \(100.0 \mathrm{km} / \mathrm{h}\) its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. (Hint: since the distance traveled is of interest rather than the time, \(x\) is the desired independent variable and not \(t .\) Use the Chain Rule to change the variable: \(\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}\)

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