/*! 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 3 Two horizontal forces, \(\overri... [FREE SOLUTION] | 91Ó°ÊÓ

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Two horizontal forces, \(\overrightarrow{\mathbf{F}}_{1}\) and \(\overrightarrow{\mathbf{F}}_{2},\) are acting on a box, but only \(\overrightarrow{\mathbf{F}}_{1}\) is shown in the drawing. \(\overrightarrow{\mathbf{F}}_{2}\) can point either to the right or to the left. The box moves only along the \(x\) axis. There is no friction between the box and the surface. Suppose that \(\overrightarrow{\mathbf{F}}_{1}=+9.0 \mathrm{N}\) and the mass of the box is \(3.0 \mathrm{kg}\). Find the magnitude and direction of \(\overrightarrow{\mathbf{F}}_{2}\) when the acceleration of the box is (a) \(+5.0 \mathrm{m} / \mathrm{s}^{2},\) (b) \(-5.0 \mathrm{m} / \mathrm{s}^{2},\) and (c) \(0 \mathrm{m} / \mathrm{s}^{2}.\)

Short Answer

Expert verified
(a) +6.0 N to the right, (b) -24.0 N to the left, (c) -9.0 N to the left.

Step by step solution

01

Apply Newton's Second Law

Newton's Second Law states that \( \overrightarrow{\mathbf{F}}_{net} = m \cdot \overrightarrow{\mathbf{a}} \), where \( \overrightarrow{\mathbf{F}}_{net} \) is the net force on the object, \( m \) is its mass, and \( \overrightarrow{\mathbf{a}} \) is its acceleration. For this problem, the net force is the sum of \( \overrightarrow{\mathbf{F}}_1 \) and \( \overrightarrow{\mathbf{F}}_2 \). Therefore, the equation becomes: \[ \overrightarrow{\mathbf{F}}_1 + \overrightarrow{\mathbf{F}}_2 = m \cdot \overrightarrow{\mathbf{a}} \] Substitute the known values of \( \overrightarrow{\mathbf{F}}_1 = +9.0 \mathrm{N} \) and mass \( m = 3.0 \mathrm{kg} \).
02

Solve for \( \overrightarrow{\mathbf{F}}_2 \) when \( a = +5.0 \mathrm{m/s}^2 \)

Substitute \( a = +5.0 \mathrm{m/s}^2 \) into the equation from Step 1: \[ 9.0 + \overrightarrow{\mathbf{F}}_2 = 3.0 \cdot 5.0 \] Simplify to find: \[ \overrightarrow{\mathbf{F}}_2 = 15.0 - 9.0 = +6.0 \mathrm{N} \] So, \( \overrightarrow{\mathbf{F}}_2 \) is \( 6.0 \mathrm{N} \) to the right.
03

Solve for \( \overrightarrow{\mathbf{F}}_2 \) when \( a = -5.0 \mathrm{m/s}^2 \)

Substitute \( a = -5.0 \mathrm{m/s}^2 \) into the equation from Step 1: \[ 9.0 + \overrightarrow{\mathbf{F}}_2 = 3.0 \cdot (-5.0) \] Simplify to find: \[ \overrightarrow{\mathbf{F}}_2 = -15.0 - 9.0 = -24.0 \mathrm{N} \] So, \( \overrightarrow{\mathbf{F}}_2 \) is \( 24.0 \mathrm{N} \) to the left.
04

Solve for \( \overrightarrow{\mathbf{F}}_2 \) when \( a = 0 \mathrm{m/s}^2 \)

Substitute \( a = 0 \mathrm{m/s}^2 \) into the equation from Step 1: \[ 9.0 + \overrightarrow{\mathbf{F}}_2 = 3.0 \cdot 0 \] This simplifies to: \[ \overrightarrow{\mathbf{F}}_2 = 0 - 9.0 = -9.0 \mathrm{N} \] Thus, \( \overrightarrow{\mathbf{F}}_2 \) is \( 9.0 \mathrm{N} \) to the left.

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Key Concepts

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

Understanding Net Force
Imagine you have multiple forces acting on an object. The net force is like the sum of all these forces combined. It tells you the overall push or pull on an object in a certain direction.
The formula for finding the net force is given by Newton's Second Law:
  • \( \overrightarrow{\mathbf{F}}_{net} = m \cdot \overrightarrow{\mathbf{a}} \)
  • Where \( \overrightarrow{\mathbf{F}}_{net} \) is the net force, \( m \) is the mass of the object, and \( \overrightarrow{\mathbf{a}} \) is the acceleration.
In the exercise, two forces are acting on a box: \( \overrightarrow{\mathbf{F}}_1 \) is known, while \( \overrightarrow{\mathbf{F}}_2 \) is what we need to find. By understanding that the net force is simply the vector sum of \( \overrightarrow{\mathbf{F}}_1 \) and \( \overrightarrow{\mathbf{F}}_2 \), you can calculate the unknown force based on the net force equation. Just remember, net force is all about the total effect of forces acting on an object at the same time.
Discovering Acceleration
Acceleration is how much the speed of an object changes over time. It's like when you press the gas pedal in a car, and the car goes faster. Acceleration can happen in any direction, depending on how the forces are applied.
In scientific terms, it's calculated using:
  • \( \overrightarrow{\mathbf{a}} = \frac{\overrightarrow{\mathbf{F}}_{net}}{m} \)
  • Where \( \overrightarrow{\mathbf{a}} \) is the acceleration, \( \overrightarrow{\mathbf{F}}_{net} \) is the net force applied, and \( m \) is the mass of the object.
In our exercise, there are three different accelerations to consider:
  • \(+5.0 \mathrm{m/s}^2\), when the object speeds up in the positive direction.
  • \(-5.0 \mathrm{m/s}^2\), when the object speeds up in the opposite direction, which means slowing down if it was originally moving in the positive direction.
  • \(0 \mathrm{m/s}^2\), indicating no change in speed.
Acceleration tells you how quickly an object reaches its speed, and it is always tied to the net force applied on it.
Grasping Force Direction
The direction of a force is super important because it tells us where the object will move or how its speed will change. In physics, we represent direction with positive or negative signs.
In this exercise, force direction plays a critical role:
  • If the force is positive, like \( \overrightarrow{\mathbf{F}}_1 = +9.0 \mathrm{N}\), it points to the right.
  • A negative force, like \( \overrightarrow{\mathbf{F}}_2 = -9.0 \mathrm{N}\), points to the left.
Understanding force direction helps in visualizing whether the forces are helping each other (in the same direction) or working against each other (in opposite directions).
For example, when \( \overrightarrow{\mathbf{F}}_2 \) opposes \( \overrightarrow{\mathbf{F}}_1 \), it can change the resulting behavior of the box, such as slowing it down when moving to the right or speeding it up to the left. The real trick is to add these vectors correctly, keeping their direction in mind, which dictates the overall movement of the object.

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Most popular questions from this chapter

An 81-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is \(0.49 .\) (a) What is the magnitude of the frictional force? (b) If the player comes to rest after \(1.6 \mathrm{s},\) what was his initial velocity?

The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force \(\overrightarrow{\mathbf{F}}_{\mathrm{sM}}\) that the sun exerts on the moon is perpendicular to the force \(\overrightarrow{\mathbf{F}}_{\mathrm{EM}}\) that the earth exerts on the moon. The masses are: mass of sun \(=1.99 \times 10^{30} \mathrm{kg},\) mass of earth \(=5.98 \times 10^{24} \mathrm{kg},\) mass of moon \(=\) \(7.35 \times 10^{22} \mathrm{kg} .\) The distances shown in the drawing are \(r_{\mathrm{SM}}=1.50 \times 10^{11} \mathrm{m}\) and \(r_{\mathrm{EM}}=3.85 \times 10^{8} \mathrm{m} .\) Determine the magnitude of the net gravitational force on the moon.

A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at \(30.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.900 , and the log has an acceleration of magnitude \(0.800 \mathrm{m} / \mathrm{s}^{2}\). Find the tension in the rope.

Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed due north. Another has a magnitude of 60.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

The speed of a bobsled is increasing because it has an acceleration of \(2.4 \mathrm{m} / \mathrm{s}^{2} .\) At a given instant in time, the forces resisting the motion, including kinetic friction and air resistance, total 450N. The combined mass of the bobsled and its riders is \(270 \mathrm{kg}\) (a) What is the magnitude of the force propelling the bobsled forward? (b) What is the magnitude of the net force that acts on the bobsled?

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