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Chapter 5: Problem 2

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 \((\mathrm{c}) \vec{F}_{2}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4.0 \mathrm{~N}) \hat{\mathrm{j}}\)

Short Answer

Expert verified
(a) \( \vec{a} = \vec{0} \); (b) \( \vec{a} = 4.0 \hat{\jmath} \); (c) \( \vec{a} = 3.0 \hat{\imath} \).

Step by step solution

01

Understanding Newton's Second Law

Newton's second law states that \( \vec{F}_{\text{net}} = m \vec{a} \), where \( \vec{F}_{\text{net}} \) is the net force acting on the object, \( m \) is the mass of the object, and \( \vec{a} \) is the acceleration. In this problem, we'll calculate the net force for each case and use it to find the acceleration.
02

Define the given forces

For each part, we have two forces: \( \vec{F}_1 = (3.0 \, \mathrm{N}) \hat{\imath} + (4.0 \, \mathrm{N}) \hat{\jmath} \). The force \( \vec{F}_2 \) varies by part. Compute the net force for each scenario.
03

Case (a): Calculate Net Force for \(\vec{F}_2 = (-3.0 \, \mathrm{N}) \hat{\imath} + (-4.0 \, \mathrm{N}) \hat{\jmath} \)

\( \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 = [(3.0 \, \mathrm{N}) + (-3.0 \, \mathrm{N})] \hat{\imath} + [(4.0 \, \mathrm{N}) + (-4.0 \, \mathrm{N})] \hat{\jmath} = 0 \hat{\imath} + 0 \hat{\jmath} \). Hence, \( \vec{F}_{\text{net}} = \vec{0} \).
04

Case (a): Calculate Acceleration

Using Newton's second law, \( \vec{F}_{\text{net}} = m \vec{a} \). Since \( \vec{F}_{\text{net}} = \vec{0} \), \( \vec{a} \) is also \( \vec{0} \).
05

Case (b): Calculate Net Force for \(\vec{F}_2 = (-3.0 \, \mathrm{N}) \hat{\imath} + (4.0 \, \mathrm{N}) \hat{\jmath} \)

\( \vec{F}_{\text{net}} = [(3.0 \, \mathrm{N}) + (-3.0 \, \mathrm{N})] \hat{\imath} + [(4.0 \, \mathrm{N}) + (4.0 \, \mathrm{N})] \hat{\jmath} = 0 \hat{\imath} + 8.0 \hat{\jmath} \).
06

Case (b): Calculate Acceleration

\( \vec{a} = \frac{\vec{F}_{\text{net}}}{m} = \frac{0 \hat{\imath} + 8.0 \hat{\jmath}}{2.0} = 0 \hat{\imath} + 4.0 \hat{\jmath} \, \mathrm{m/s^2} \).
07

Case (c): Calculate Net Force for \(\vec{F}_2 = (3.0 \, \mathrm{N}) \hat{\imath} + (-4.0 \, \mathrm{N}) \hat{\jmath} \)

\( \vec{F}_{\text{net}} = [(3.0 \, \mathrm{N}) + (3.0 \, \mathrm{N})] \hat{\imath} + [(4.0 \, \mathrm{N}) + (-4.0 \, \mathrm{N})] \hat{\jmath} = 6.0 \hat{\imath} + 0 \hat{\jmath} \).
08

Case (c): Calculate Acceleration

\( \vec{a} = \frac{\vec{F}_{\text{net}}}{m} = \frac{6.0 \hat{\imath} + 0 \hat{\jmath}}{2.0} = 3.0 \hat{\imath} + 0 \hat{\jmath} \, \mathrm{m/s^2} \).

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Key Concepts

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

Vector Addition
When dealing with forces, vectors are a useful way to represent them. Vectors have both magnitude and direction, which makes them perfect for describing things like forces.
Even though it's tempting to handle the components separately, remember that
  • Vectors have to be added or subtracted in their entirety, accounting for both magnitude and direction.
  • In two-dimensional space, each vector can be broken down into components, often represented as the i-component (x-direction) and j-component (y-direction).

To perform vector addition with forces, add the respective components:
  • i-components (x-direction): Add the i-component of the first force to the i-component of the second force.
  • j-components (y-direction): Add the j-component of the first force to the j-component of the second force.

This method helps you find the resultant force or net force acting on an object. Understanding vector addition ensures that the directions of the forces are accurately accounted for, which is crucial for determining the correct net force.
Net Force
The net force is essentially the sum of all forces acting on an object. According to Newton's Second Law \(\vec{F}_{\text{net}} = m \vec{a}\), where \(\vec{F}_{\text{net}}\) represents the net force, \(m\) is mass, and \(\vec{a}\) is acceleration.
  • Identifying the net force involves adding all acting forces as vectors.
  • It's not just about sum of magnitudes, directions matter a lot!
  • If forces cancel each other, the net force can be zero!

In the given exercise, understanding the signs (+/-) is important:
  • If forces are in opposite directions, their net result might reduce or may cancel out.
  • For example, in part (a) of the solution, the two forces exactly cancel each other out, resulting in no net force.
Reasoning about how forces and their directions combine leads us to crucial insights into their effects: consistent contributions (like in the y-direction of part (b)) reinforce each other, while opposing contributions (like in the y-direction of part (a)) diminish them.
Acceleration Calculation
Acceleration is a measure of how fast velocity changes, and it's directly tied to the net force acting on an object through Newton's Second Law.
To compute acceleration:
  • Use the formula \(\vec{a} = \frac{\vec{F}_{\text{net}}}{m}\).
  • Calculate the net force first using vector addition.
  • Divide each component of \(\vec{F}_{\text{net}}\) by the mass \(m\).

In this exercise, given a chopping block's mass of \(2.0\, \mathrm{kg}\), you were provided different force scenarios:
  • In scenario (a), the net force was zero, hence, no change in motion reflecting \(\vec{a} = \vec{0}\).
  • In scenario (b), the net force in the y-direction led to \(\vec{a} = 8.0/2.0 = 4.0\, \mathrm{m/s^2}\) in the j-component.
  • In scenario (c), it led to acceleration \(\vec{a} = 3.0\, \mathrm{m/s^2}\) solely in the i-direction.

Understanding acceleration in terms of individual components is key, as it tells us how the object moves in each dimension separately. This clarity simplifies how effects like stopping and changing direction are predicted.

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Most popular questions from this chapter

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m}\). He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

A certain particle has a weight of \(22 \mathrm{~N}\) at a point where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\). What are its (a) weight and (b) mass at a point where \(g=4.9 \mathrm{~m} / \mathrm{s}^{2} ?\) What are its (c) weight and (d) mass if it is moved to a point in space where \(g=0\) ?

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times\) \(10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg}\). Determine the vertical distance the electron is deflected during the time it has moved \(30 \mathrm{~mm}\) horizontally.

Sunjamming. 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?

For sport, a \(12 \mathrm{~kg}\) armadillo runs onto a large pond of level, frictionless ice. The armadillo's initial velocity is \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis. Take its initial position on the ice as being the origin. It slips over the ice while being pushed by a wind with a force of \(17 \mathrm{~N}\) in the positive direction of the \(y\) axis. In unitvector notation, what are the animal's (a) velocity and (b) position vector when it has slid for \(3.0\) s?
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