/*! 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 2 Two horizontal forces act on a \... [FREE SOLUTION] | 91Ó°ÊÓ

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Two horizontal forces act on a \(2.0 \mathrm{~kg}\) chopping block that can slide over a frictionless kitchen counter, which lies in an \(x y\) plane. One force is \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\). Find the acceleration of the chopping block in unit-vector notation when the other force is (a) \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4,0 \mathrm{~N}) \hat{\mathrm{j}}\), (b) \(\vec{F}_{2}=(-3,0 \mathrm{~N}) \hat{\mathrm{i}}+(4,0 \mathrm{~N}) \hat{\mathrm{j}}\), and (c) \(\vec{F}_{2}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4.0 \mathrm{~N}) \hat{\mathrm{j}}\)

Short Answer

Expert verified
(a) \( 0 \mathrm{~m/s^2} \), (b) \( 4.0 \mathrm{~m/s^2} \hat{\mathrm{j}} \), (c) \( 3.0 \mathrm{~m/s^2} \hat{\mathrm{i}} \)

Step by step solution

01

Identify Given Values

The mass of the chopping block is given as \( m = 2.0 \text{ kg} \). The force \( \vec{F}_1 \) is \( (3.0 \mathrm{~N}) \hat{\mathrm{i}} + (4.0 \mathrm{~N}) \hat{\mathrm{j}} \). We need to find the acceleration when different \( \vec{F}_2 \) forces are applied.
02

Apply Newton's Second Law

Newton's Second Law states \( \vec{a} = \frac{\vec{F}_\text{net}}{m} \), where \( \vec{a} \) is the acceleration, \( \vec{F}_\text{net} \) is the net force, and \( m \) is the mass. We will apply this law to find the acceleration for each case.
03

Calculate Net Force for Case (a)

In case (a), \( \vec{F}_2 = (-3.0 \mathrm{~N}) \hat{\mathrm{i}} + (-4.0 \mathrm{~N}) \hat{\mathrm{j}} \). The net force \( \vec{F}_\text{net} = \vec{F}_1 + \vec{F}_2 = (3.0 \mathrm{~N} + (-3.0 \mathrm{~N})) \hat{\mathrm{i}} + (4.0 \mathrm{~N} + (-4.0 \mathrm{~N})) \hat{\mathrm{j}} = (0 \mathrm{~N}) \hat{\mathrm{i}} + (0 \mathrm{~N}) \hat{\mathrm{j}} \).
04

Find Acceleration for Case (a)

Using \( \vec{a} = \frac{\vec{F}_\text{net}}{m} \), the acceleration \( \vec{a} = \frac{(0 \mathrm{~N}) \hat{\mathrm{i}} + (0 \mathrm{~N}) \hat{\mathrm{j}}}{2.0 \mathrm{~kg}} = (0 \mathrm{~m/s^2}) \hat{\mathrm{i}} + (0 \mathrm{~m/s^2}) \hat{\mathrm{j}} \).
05

Calculate Net Force for Case (b)

In case (b), \( \vec{F}_2 = (-3.0 \mathrm{~N}) \hat{\mathrm{i}} + (4.0 \mathrm{~N}) \hat{\mathrm{j}} \). The net force \( \vec{F}_\text{net} = \vec{F}_1 + \vec{F}_2 = (3.0 \mathrm{~N} + (-3.0 \mathrm{~N})) \hat{\mathrm{i}} + (4.0 \mathrm{~N} + 4.0 \mathrm{~N}) \hat{\mathrm{j}} = (0 \mathrm{~N}) \hat{\mathrm{i}} + (8.0 \mathrm{~N}) \hat{\mathrm{j}} \).
06

Find Acceleration for Case (b)

Using \( \vec{a} = \frac{\vec{F}_\text{net}}{m} \), the acceleration \( \vec{a} = \frac{(0 \mathrm{~N}) \hat{\mathrm{i}} + (8.0 \mathrm{~N}) \hat{\mathrm{j}}}{2.0 \mathrm{~kg}} = (0 \mathrm{~m/s^2}) \hat{\mathrm{i}} + (4.0 \mathrm{~m/s^2}) \hat{\mathrm{j}} \).
07

Calculate Net Force for Case (c)

In case (c), \( \vec{F}_2 = (3.0 \mathrm{~N}) \hat{\mathrm{i}} + (-4.0 \mathrm{~N}) \hat{\mathrm{j}} \). The net force \( \vec{F}_\text{net} = \vec{F}_1 + \vec{F}_2 = (3.0 \mathrm{~N} + 3.0 \mathrm{~N}) \hat{\mathrm{i}} + (4.0 \mathrm{~N} + (-4.0 \mathrm{~N})) \hat{\mathrm{j}} = (6.0 \mathrm{~N}) \hat{\mathrm{i}} + (0 \mathrm{~N}) \hat{\mathrm{j}} \).
08

Find Acceleration for Case (c)

Using \( \vec{a} = \frac{\vec{F}_\text{net}}{m} \), the acceleration \( \vec{a} = \frac{(6.0 \mathrm{~N}) \hat{\mathrm{i}} + (0 \mathrm{~N}) \hat{\mathrm{j}}}{2.0 \mathrm{~kg}} = (3.0 \mathrm{~m/s^2}) \hat{\mathrm{i}} + (0 \mathrm{~m/s^2}) \hat{\mathrm{j}} \).

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Key Concepts

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

Net Force Calculation
Calculating the net force acting on an object is an essential step in applying Newton's Second Law. To find the net force, combine all the individual forces acting on the object.
Forces are vector quantities, which means they have both a magnitude and a direction. When dealing with forces in problems like this, you'll often encounter them represented as vectors in a coordinate system.
  • The net force (\( \vec{F}_\text{net} \)) is determined by vector addition of all acting forces.
  • This ensures you take into account each force's direction while summing their magnitudes.
In the given problem, we are provided with two forces, \( \vec{F}_1 \) and \( \vec{F}_2 \), that act on a chopping block. By adding these two vectors together, we find the net force in each scenario.
For example, in Case (a), the forces \( \vec{F}_1 = (3.0 \mathrm{~N}) \hat{\mathrm{i}} + (4.0 \mathrm{~N}) \hat{\mathrm{j}} \) and \( \vec{F}_2 = (-3.0 \mathrm{~N}) \hat{\mathrm{i}} + (-4.0 \mathrm{~N}) \hat{\mathrm{j}} \) are added to find \( \vec{F}_\text{net} = (0 \mathrm{~N}) \hat{\mathrm{i}} + (0 \mathrm{~N}) \hat{\mathrm{j}} \).
Each case follows a similar process to ensure the net force is calculated accurately.
Vector Addition
Understanding vector addition is crucial to solving problems involving forces in multiple directions. In most physical situations, forces do not always act neatly along one axis. Hence, you need to combine them to find the resultant force.
Vector addition involves two main components:
  • Add vectors that are aligned along the same direction directly to get a total.
  • Separately add their i-components (horizontal) and j-components (vertical) to find the net vector's magnitude and direction.
In our exercise, each force has two components: i-components along the x-axis and j-components along the y-axis. When dealing with vectors like \( \vec{F}_1 \) and \( \vec{F}_2 \), follow these steps:
  • Align each vector's tail to the origin in the coordinate plane.
  • Add up \( \hat{\mathrm{i}} \) components by summing their respective coefficients.
  • Add up \( \hat{\mathrm{j}} \) components similarly.
  • Resultant is the summation vector, which gives both direction and magnitude of your net force.
This approach helps create a clearer picture of how forces balance or cause motion in any direction.
Acceleration Calculation
After determining the net force, calculating the acceleration becomes straightforward using Newton's Second Law. This law articulates that the acceleration (\( \vec{a} \)) of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula used here is:\[ \vec{a} = \frac{\vec{F}_\text{net}}{m} \]Where:
  • \( \vec{F}_\text{net} \) is the net force vector.
  • \( m \) is the mass of the object (in kg).
In our problem, the chopping block has a mass of \( 2.0 \text{ kg} \). For each net force calculation in the exercise, substitute \( \vec{F}_\text{net} \) and the mass into the formula to find acceleration.
For instance, in Case (b), the net force is \( (0 \mathrm{~N}) \hat{\mathrm{i}} + (8.0 \mathrm{~N}) \hat{\mathrm{j}} \) and thus, the acceleration is calculated as:\[ \vec{a} = \frac{(0 \mathrm{~N}) \hat{\mathrm{i}} + (8.0 \mathrm{~N}) \hat{\mathrm{j}}}{2.0 \mathrm{~kg}} = (0 \mathrm{~m/s^2}) \hat{\mathrm{i}} + (4.0 \mathrm{~m/s^2}) \hat{\mathrm{j}} \]This method shows how forces result in acceleration, important especially in physics problems involving motion and dynamics.

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Most popular questions from this chapter

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13\) \(\mathrm{g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s}\), and its penetration depth was \(15 \mathrm{~mm}\). If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

A shot putter launches a \(7.260 \mathrm{~kg}\) shot by pushing it along a straight line of length \(1.650 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\) from the horizontal, accelerating the shot to the launch speed from its initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of \(2.110\) \(\mathrm{m}\) and at an angle of \(34.10^{\circ}\), and it lands at a horizontal distance of \(15.90 \mathrm{~m}\). What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)

A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of \(900 \mathrm{~kg}\) and receives a push of \(20 \mathrm{~N}\). (a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

A car that weighs \(1.30 \times 10^{4} \mathrm{~N}\) is initially moving at \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m}\). Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

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