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Chapter 8: Problem 3

A \(2.0\) -kg brick is moving at a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). How large a force \(F\) is needed to stop the brick in a time of \(7.0 \times 10^{-4} \mathrm{~s}\) ? Since we have a force and the time over which it acts, that suggests using the impulse equation (i.e., Newton's Second Law): Impulse on brick \(=\) Change in momentum of brick $$ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F\left(7.0 \times 10^{-4} \mathrm{~s}\right) &=0-(2.0 \mathrm{~kg})(6.0 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ from which \(F=-1.7 \times 10^{4} \mathrm{~N}\). The minus sign indicates that the force opposes the motion.

Short Answer

Expert verified
The force needed is approximately \(-1.7 \times 10^4 \, \text{N}\), indicating it opposes the motion.

Step by step solution

01

Understand the Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse on an object is equal to the change in its momentum. Mathematically, it is expressed as: \[ F \Delta t = \Delta p \] where \( F \) is the force, \( \Delta t \) is the time period over which the force acts, and \( \Delta p \) is the change in momentum.
02

Identify Initial and Final Conditions

Identify the initial momentum \( p_i = m v_i \) and final momentum \( p_f = m v_f \). Given data:- Mass \( m = 2.0 \, \text{kg} \)- Initial velocity \( v_i = 6.0 \, \text{m/s} \)- Final velocity \( v_f = 0 \, \text{m/s} \).Calculate initial momentum: \( p_i = 2.0 \, \text{kg} \times 6.0 \, \text{m/s} = 12.0 \, \text{kg m/s} \).Final momentum: \( p_f = 2.0 \, \text{kg} \times 0 = 0 \, \text{kg m/s} \).
03

Calculate Change in Momentum

Calculate the change in momentum \( \Delta p = p_f - p_i \):\[ \Delta p = 0 - 12.0 \, \text{kg m/s} = -12.0 \, \text{kg m/s} \]The negative sign indicates the change is in the opposite direction of initial motion.
04

Apply Impulse-Momentum Theorem to Solve for Force

Use the impulse equation \( F \Delta t = \Delta p \) to find the force:- Time \( \Delta t = 7.0 \times 10^{-4} \, \text{s} \)Substitute the known values:\[ F \times 7.0 \times 10^{-4} \, \text{s} = -12.0 \, \text{kg m/s} \]Solve for \( F \):\[ F = \frac{-12.0 \, \text{kg m/s}}{7.0 \times 10^{-4} \, \text{s}} \approx -1.7 \times 10^4 \, \text{N} \]
05

Interpret the Result

The calculated force \( F \approx -1.7 \times 10^4 \, \text{N} \). The negative sign indicates that the force acts in the opposite direction to the motion to stop the brick.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle of physics that relates the motion of an object to the forces acting on it. It is expressed by the equation \( \mathbf{F} = m \mathbf{a} \). Here, \( F \) represents the net force acting on an object, \( m \) is the object's mass, and \( a \) is its acceleration. Imagine pushing a brick across a smooth surface. Depending on how hard you push, the brick will accelerate more or less. That's Newton's Second Law at work.
  • A larger force results in greater acceleration when the mass stays constant.
  • More massive objects require a greater force to achieve the same acceleration as lighter objects.
When applying the Impulse-Momentum Theorem, Newton's Second Law aids in understanding how force affects the change in velocity, and thus momentum, over time.
Momentum Change
Momentum is a measure of an object's motion and is calculated as the product of an object's mass and velocity, expressed as \( p = mv \). When an object speeds up, slows down, or changes direction, its momentum changes. The rate of this change is particularly important in calculating the effects of forces over time.
  • Initial momentum, \( p_i \), can be calculated with the initial velocity.
  • Final momentum, \( p_f \), relates to the state where the object stops or changes speed.
In the case of the brick problem, the initial momentum can be calculated as \( 2.0 \, \text{kg} \times 6.0 \, \text{m/s} = 12.0 \, \text{kg m/s} \). The final momentum being zero reflects the brick having stopped.
Force Calculation
Determining the force required to change an object's momentum over a specified time involves the Impulse-Momentum Theorem. Here the impulse, the product of force \( F \) and time \( \Delta t \), is set equal to the change in momentum \( \Delta p \). This is expressed as:\[ F \Delta t = \Delta p \]Substituting in the values for the brick,
  • The change in momentum is \( -12.0 \, \text{kg m/s} \).
  • The time over which this change occurs is \( 7.0 \times 10^{-4} \, \text{s} \).
Solving for the force \( F \) gives:\[ F = \frac{-12.0 \, \text{kg m/s}}{7.0 \times 10^{-4} \, \text{s}} = -1.7 \times 10^4 \, \text{N} \] The negative sign indicates that the direction of the force is opposite to the initial motion, meaning it acts to stop the brick.
Initial and Final Conditions
Identifying initial and final conditions helps establish the parameters for solving the problem. Initial conditions include the object's initial velocity and momentum, while final conditions provide the desired end state of the object. For the scenario with the brick, clarity is achieved by noting:
  • Initial velocity \( v_i = 6.0 \, \text{m/s} \).
  • Final velocity \( v_f = 0 \, \text{m/s} \), since the goal is to stop the brick.
  • Initial momentum \( p_i = 12.0 \, \text{kg m/s} \).
  • Final momentum \( p_f = 0 \, \text{kg m/s} \).
These conditions allow us to calculate the change in momentum and subsequently the force needed to achieve this transformation.

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Most popular questions from this chapter

An empty \(15000-\mathrm{kg}\) coal car is coasting on a level track at \(5.00 \mathrm{~m} / \mathrm{s}\). Suddenly \(5000 \mathrm{~kg}\) of coal is dumped into it from directly above it. The coal initially has zero horizontal velocity with respect to the ground. Find the final speed of the car.

Two identical balls undergo a collision at the origin of coordinates. Before collision their scalar velocity components are \(\left(u_{x}=40 \mathrm{~cm} / \mathrm{s}, u_{y}=0\right)\) and \(\left(u_{x}=-30 \mathrm{~cm} / \mathrm{s}, u_{y}=20 \mathrm{~cm} / \mathrm{s}\right) .\) After collision, the first ball (the one moving along the \(x\) -axis) is standing still. Find the scalar velocity components of the second ball. [Hint: After the collision, the moving ball must have all of the momentum of the system.]

An \(8.0\) -g bullet is fired horizontally into a \(9.00\) -kg cube of wood, which is at rest on a frictionless air table. The bullet lodges in the wood. The cube is free to move and has a speed of \(40 \mathrm{~cm} / \mathrm{s}\) after impact. Find the initial velocity of the bullet. This is an example of a completely inelastic collision for which momentum is conserved, although \(\mathrm{KE}\) is not. Consider the system (cube \(+\) bullet). The velocity, and hence the momentum, of the cube before impact is zero. Take the bullet's initial motion to be positive in the positive \(x\) -direction. The momentum conservation law tells us that Momentum of system before impact = Momentum of system after impact $$ \begin{aligned} \text { (Momentum of bullet) }+\text { (Momentum of cube) } &=(\text { Momentum of bullet }+\text { Cube }) \\ m_{B} v_{B x}+m_{C} v_{C x} &=\left(m_{B}+m_{C}\right) v_{x} \\ (0.0080 \mathrm{~kg}) v_{B x}+0 &=(9.008 \mathrm{~kg})(0.40 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ Solving yields \(v_{B x}=0.45 \mathrm{~km} / \mathrm{s}\) and so \(\overrightarrow{\mathbf{v}}_{B}=0.45 \mathrm{~km} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION.

What average resisting force must act on a \(3.0-\mathrm{kg}\) mass to reduce its speed from \(65 \mathrm{~cm} / \mathrm{s}\) to \(15 \mathrm{~cm} / \mathrm{s}\) in \(0.20 \mathrm{~s}\) ?

Four masses are positioned in the \(x y\) -plane as follows: \(300 \mathrm{~g}\) at \((x=0, y=2.0 \mathrm{~m}), 500 \mathrm{~g}\) at \((-20 \mathrm{~m},-3.0 \mathrm{~m})\), \(700 \mathrm{~g}\) at \((50 \mathrm{~cm}, 30 \mathrm{~cm})\), and \(900 \mathrm{~g}\) at \((-80 \mathrm{~cm}, 150 \mathrm{~cm})\). Find their center of mass.
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