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Chapter 4: Problem 113

A duck has a mass of \(2.5 \mathrm{kg}\). As the duck paddles, a force of \(0.10 \mathrm{N}\) acts on it in a direction due east. In addition, the current of the water exerts a force of \(0.20 \mathrm{N}\) in a direction of \(52^{\circ}\) south of east. When these forces begin to act, the velocity of the duck is \(0.11 \mathrm{m} / \mathrm{s}\) in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in \(3.0 \mathrm{s}\) while the forces are acting.

Short Answer

Expert verified
The duck displaces approximately 0.78 m at an angle of 21.4掳 south of east.

Step by step solution

01

Determine the net force components

First, we need to resolve the forces into their components. The force due east is \(0.10\,\mathrm{N}\), and does not need to resolve into components because it acts entirely along the x-axis. Next, resolve the \(0.20\,\mathrm{N}\) force into x and y components using trigonometry:- The x-component of \(0.20\,\mathrm{N}\) is \(0.20 \cos(52^{\circ})\).- The y-component of \(0.20\,\mathrm{N}\) is \(0.20 \sin(52^{\circ})\). This is negative as it's acting south.Calculate:\[ F_{x,\,\text{total}} = 0.10 + 0.20 \cos(52^{\circ}) \approx 0.22\,\mathrm{N} \]\[ F_{y,\,\text{total}} = -0.20 \sin(52^{\circ}) \approx -0.16\,\mathrm{N} \]
02

Calculate the net acceleration

Use Newton's second law \(F = ma\) to find the acceleration components. The duck鈥檚 mass \(m\) is \(2.5\,\mathrm{kg}\):\[ a_{x} = \frac{F_{x,\,\text{total}}}{m} = \frac{0.22}{2.5} = 0.088\,\mathrm{m/s^2} \]\[ a_{y} = \frac{F_{y,\,\text{total}}}{m} = \frac{-0.16}{2.5} = -0.064\,\mathrm{m/s^2} \]
03

Calculate the change in velocity components

Using the acceleration components found and time \(t = 3.0\,\mathrm{s}\), calculate the change in velocity:\[ \Delta v_{x} = a_{x} \cdot t = 0.088 \cdot 3.0 = 0.264\,\mathrm{m/s} \]\[ \Delta v_{y} = a_{y} \cdot t = -0.064 \cdot 3.0 = -0.192\,\mathrm{m/s} \]
04

Determine the final velocity components

Add the change in velocity to the initial components. The initial velocity is given as \(0.11\,\mathrm{m/s}\) due east (positive x direction):\[ v_{x, \text{final}} = 0.11 + 0.264 = 0.374\,\mathrm{m/s} \]\[ v_{y, \text{final}} = -0.192\,\mathrm{m/s} \]
05

Calculate the displacement components

Use the formula \(\Delta x = v_{x,0} t + \frac{1}{2} a_{x} t^2\) and similarly for \(\Delta y\). Here \(v_{x,0} = 0.11\,\mathrm{m/s}\):\[ \Delta x = 0.11 \cdot 3 + \frac{1}{2} \cdot 0.088 \cdot 3^2 = 0.33 + 0.396 = 0.726\,\mathrm{m} \]\[ \Delta y = 0 + \frac{1}{2} \cdot (-0.064) \cdot 3^2 = -0.288\,\mathrm{m} \]
06

Calculate magnitude and direction of displacement

The magnitude of the displacement is calculated using the Pythagorean theorem:\[ \Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(0.726)^2 + (-0.288)^2} \approx 0.78\,\mathrm{m} \]The direction \(\theta\) relative to the east (x-axis) can be found using:\[ \theta = \tan^{-1}\left(\frac{\Delta y}{\Delta x}\right) = \tan^{-1}\left(\frac{-0.288}{0.726}\right) \approx -21.4^{\circ} \]This angle is measured south of east.

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Key Concepts

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

Understanding Force Components
To analyze any system involving forces acting at angles, breaking them into components along perpendicular axes is essential. This simplifies the calculation of net forces drastically. In our scenario, we have two forces acting on the duck: a 0.10 N force due east and a 0.20 N force at 52掳 south of east.
Breaking the angled force into components, we use trigonometry:
  • The x-component (horizontal) is found using the cosine of the angle: \[ F_{x} = 0.20 imes \cos(52^{\circ}) \]
  • The y-component (vertical) is obtained using the sine of the angle: \[ F_{y} = 0.20 imes \sin(52^{\circ}) \]
Remember, the y-component is negative because it acts south, which is conventionally considered down or negative in the coordinate plane. Accurate component resolution is critical for correctly determining subsequent physical quantities like acceleration and displacement.
Calculating the Net Force
Net force is the vector sum of all individual forces acting on an object. For the duck, we need to combine the forces acting in both the x and y directions.
The total force due east (x-direction) is:
  • Including the 0.10 N force, resulting in:
    • \[ F_{x,\text{total}} = 0.10 + 0.20 \cos(52^{\circ}) \]
In the y-direction, where only the current acts:
  • The total force is:\[ F_{y,\text{total}} = -0.20 \sin(52^{\circ}) \]
These components help us determine the net force vector, crucial for understanding how the duck will accelerate. Calculating net force correctly forms a basis for applying Newton's second law to find the resulting acceleration.
Acceleration Calculation
Once the net force components are known, Newton's second law \( F = ma \) allows us to calculate the acceleration of an object. For the duck, with a mass of 2.5 kg:
  • In the x-direction, the acceleration \( a_{x} \) is:\[ a_{x} = \frac{F_{x,\text{total}}}{m} \]
  • In the y-direction, the acceleration \( a_{y} \) is:\[ a_{y} = \frac{F_{y,\text{total}}}{m} \]
These equations provide the rate of change of velocity for the duck in both directions. Recognizing that acceleration directly furthers our understanding of how quickly and in what direction the duck's speed changes is crucial. This helps predict future position changes, leading us smoothly toward displacement calculations.
Displacement Calculation
With acceleration components known, determining how far and in what direction the duck moves over time involves kinematic equations. Given the initial velocity and acceleration, we use the displacement formulas:
  • For x-displacement \( \Delta x \):\[ \Delta x = v_{x,0} \cdot t + \frac{1}{2} \cdot a_{x} \cdot t^2 \]
  • For y-displacement \( \Delta y \):\[ \Delta y = v_{y,0} \cdot t + \frac{1}{2} \cdot a_{y} \cdot t^2 \]
The variables \( v_{x,0} \) and \( v_{y,0} \) are the initial velocities in each direction. In this problem, initialize \( v_{y,0} \) at zero since the duck initially travels due east. Over 3 seconds, we predict the path's change and then calculate the overall displacement with the Pythagorean theorem. Understanding displacement is vital for realizing distance and direction changes due to complex force interactions over time.

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Most popular questions from this chapter

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