/*! 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 20 (ao In Fig. 6-27, a box of Cheer... [FREE SOLUTION] | 91Ó°ÊÓ

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(ao In Fig. 6-27, a box of Cheerios (mass \(m_{C}=1.0 \mathrm{~kg}\) ) and a box of Wheaties (mass \(m_{W}=3.0\) \(\mathrm{kg}\) ) are accelerated across a horizontal surface by a horizontal force \(\quad\) Figure 6-27 Problem 20 . \(\vec{F}\) applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is \(2.0 \mathrm{~N}\), and the magnitude of the frictional force on the Wheaties box is \(4.0 \mathrm{~N}\). If the magnitude of \(\vec{F}\) is \(12 \mathrm{~N}\), what is the magnitude of the force on the Wheaties box from the Cheerios box?

Short Answer

Expert verified
The force on the Wheaties box from the Cheerios box is 8.5 N.

Step by step solution

01

Understand the problem

We have two boxes, a Cheerios box with mass \(m_C = 1.0\, \text{kg}\) and a Wheaties box with mass \(m_W = 3.0\, \text{kg}\). A force \(\vec{F} = 12\,\text{N}\) is applied to the Cheerios box. There is a frictional force of \(2.0\, \text{N}\) on the Cheerios box and \(4.0\, \text{N}\) on the Wheaties box. We need to find the force exerted by the Cheerios box on the Wheaties box.
02

Calculate total mass and net force

Calculate the total mass of the two boxes: \(m = m_C + m_W = 1.0\, \text{kg} + 3.0\, \text{kg} = 4.0\, \text{kg}\). The total friction force is \(2.0\, \text{N} + 4.0\, \text{N} = 6.0\, \text{N}\). The net force leading to acceleration is \(F_{net} = 12\, \text{N} - 6.0\, \text{N} = 6.0\, \text{N}\).
03

Calculate acceleration of the system

Use Newton’s second law \( F_{net} = m \cdot a \) to find acceleration. Solving for acceleration, \[ a = \frac{F_{net}}{m} = \frac{6.0\, \text{N}}{4.0\, \text{kg}} = 1.5\, \text{m/s}^2 \].
04

Calculate the force on Wheaties box

Now, apply Newton's second law to the Wheaties box only. The force on it by the Cheerios box minus the friction force gives its net force: \(F_{W} - 4.0\, \text{N} = 3.0\, \text{kg} \cdot 1.5\, \text{m/s}^2\).
05

Solve for force from Cheerios to Wheaties box

Solve for \(F_{W}\): \[ F_{W} = 3.0\, \text{kg} \cdot 1.5\, \text{m/s}^2 + 4.0\, \text{N} = 4.5\, \text{N} + 4.0\, \text{N} = 8.5\, \text{N} \].

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Key Concepts

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

Understanding Frictional Force
Frictional force plays a crucial role in many everyday activities, especially when it comes to objects in motion. In physics, it is the force that opposes the motion of an object, keeping it from sliding uncontrollably. This force occurs at the surface where two objects come in contact.
\[ \text{Frictional Force} = \mu \times N \]
Here, \( \mu \) is the coefficient of friction, and \( N \) is the normal force, which is usually equivalent to the weight of the object when on a flat surface.
In the case of our problem, two frictional forces are acting on the boxes. The Cheerios box experiences a frictional force of \(2.0\, \text{N}\), while the Wheaties box experiences \(4.0\, \text{N}\) of frictional resistance. This friction opposes the motion caused by the applied force \( \vec{F} \). Without these forces, the boxes would require less force to maintain the same acceleration.
Understanding these frictional influences helps us to accurately calculate the net force acting on the system.
Understanding Net Force
The concept of net force is central to solving problems involving motion. It is essentially the sum of all the forces acting on an object. According to Newton's second law, an object's acceleration is directly proportional to the net force acting upon it and inversely proportional to its mass.
\[ F_{net} = m \cdot a \]
In our exercise, the net force is calculated by subtracting the total frictional force from the applied force. Here’s how it breaks down:
  • Total applied force \(= 12.0\, \text{N}\)
  • Total frictional force \(= 2.0\, \text{N} + 4.0\, \text{N} = 6.0\, \text{N}\)
  • Therefore, \( F_{net} = 12.0\, \text{N} - 6.0\, \text{N} = 6.0\, \text{N}\)
This net force is what accelerates the entire system of boxes. It's important to understand that the net force isn't just a simple sum—it's the result of all forces adding up, both in the direction of the primary force and opposing it, such as friction.
Understanding Acceleration
Acceleration is a key concept in Newton's laws, describing how an object's velocity changes over time due to applied forces. The net force acting on an object, according to Newton's second law, leads to its acceleration.
To calculate acceleration, use the formula:
  • \( a = \frac{F_{net}}{m} \)

Substituting the values from the exercise:
  • Total mass of the boxes \( = 4.0\, \text{kg}\)
  • Net force \( = 6.0\, \text{N}\)
  • \( a = \frac{6.0\, \text{N}}{4.0\, \text{kg}} = 1.5\, \text{m/s}^{2} \)
This acceleration tells us how quickly the velocity of the boxes is increasing as they move across the floor. Knowing this rate of acceleration allows us to solve for other forces at play, like the force exerted by the Cheerios box on the Wheaties box. It's the key to understanding the dynamics within this particular system.

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Most popular questions from this chapter

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle \(200 \mathrm{~m}\) slope in the Alps, a skier reduced his top-to-bottom time from \(61 \mathrm{~s}\) with standard skis to \(42 \mathrm{~s}\) with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a \(3.0^{\circ}\) slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

A \(3.5 \mathrm{~kg}\) block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

An airplane is flying in a horizontal circle at a speed of \(480 \mathrm{~km} / \mathrm{h}\) (Fig. 6-41). If its wings are tilted at angle \(\theta=40^{\circ}\) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the \(3.5 \mathrm{~kg}\) book is pushed from rest through a distance of \(0.90 \mathrm{~m}\) by the horizontal \(25 \mathrm{~N}\) force from the broom and then has a speed of \(1.60 \mathrm{~m} / \mathrm{s}\), what is the coefficient of kinetic friction between the book and floor?

ao A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. \(6-17\) ). The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N}\),(b) \(10 \mathrm{~N}\), and \((\mathrm{c}) 12 \mathrm{~N}\)

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