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Chapter 4: Problem 12

At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or \(x\) component of the ball's acceleration is \(810 \mathrm{m} / \mathrm{s}^{2}\) and the vertical or \(y\) component of its acceleration is \(1100 \mathrm{m} / \mathrm{s}^{2} .\) The ball's mass is \(0.43 \mathrm{kg} .\) What is the magnitude of the net force acting on the soccer ball at this instant?

Short Answer

Expert verified
The magnitude of the net force is approximately 587.48 N.

Step by step solution

01

Understand the Components of Force

To find the net force on the ball, remember that force is a vector and it acts in directions. The given problem provides the acceleration in two directions: horizontal (x-component) is \( a_x = 810 \ \text{m/s}^2 \) and vertical (y-component) is \( a_y = 1100 \ \text{m/s}^2 \).
02

Use Newton's Second Law

According to Newton's Second Law, the force acting on an object is given by \( F = ma \), where \( m \) is the mass and \( a \) is the acceleration. Since there are two components of acceleration, calculate the components of force: \( F_x = m \cdot a_x = 0.43 \cdot 810 \) and \( F_y = m \cdot a_y = 0.43 \cdot 1100 \).
03

Calculate the Components of Force

Calculate each component using the mass of the ball. For the x-component: \( F_x = 0.43 \times 810 = 348.3 \ \text{N} \). For the y-component: \( F_y = 0.43 \times 1100 = 473 \ \text{N} \).
04

Find the Magnitude of the Net Force

The net force is the vector sum of the two components, calculated using the Pythagorean theorem. Thus, the magnitude \( F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{348.3^2 + 473^2} \).
05

Final Calculation for the Magnitude

Compute the above expression: \( F_{net} = \sqrt{348.3^2 + 473^2} = \sqrt{121378.89 + 223729} = \sqrt{345107.89} \approx 587.48 \ \text{N} \). Thus, the magnitude of the net force is approximately \( 587.48 \ \text{N} \).

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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 cornerstone of classical mechanics and explains how the motion of an object changes when it is subjected to forces. The law is often written in the form of the equation:
  • \( F = m imes a \),where \( F \) is the net force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration.
This law highlights that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. It essentially tells us how a force applied to an object influences its motion.
For example, kicking a soccer ball applies a force that results in the ball's acceleration. By understanding the values for mass and acceleration components, we can compute the net force acting on the ball.
Vector Components
In physics, vectors are used to represent forces that have both a magnitude and a direction. Every vector can be broken down into its components, commonly along the horizontal (x-axis) and vertical (y-axis) directions. This technique is especially useful when dealing with forces like in this soccer ball example.
  • The problem provides us with two components of acceleration: the horizontal (x-component), which is \( a_x = 810 \text{ m/s}^2 \), and the vertical (y-component), which is \( a_y = 1100 \text{ m/s}^2 \).
  • To find the net force, we first calculate the force in each direction using \( F = m \times a \):
    • For the x-component, \( F_x = 0.43 \times 810 = 348.3 \text{ N} \).
    • For the y-component, \( F_y = 0.43 \times 1100 = 473 \text{ N} \).
By breaking down the vector into components, it becomes easier to apply mathematical techniques such as the Pythagorean theorem to find the resulting net force.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle used in geometry, and it can be applied in physics to calculate the resultant magnitude of vector forces. This theorem states that in a right-angled triangle, the square of the length of the hypotenuse (longest side) is equal to the sum of the squares of the lengths of the other two sides.
  • In formula terms, this is written as \( c^2 = a^2 + b^2 \).
When calculating the net force on the soccer ball, think of the x and y force components as the two sides of a right triangle. The net force is like the hypotenuse:
  • Using the values we calculated: \( F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{348.3^2 + 473^2} \).
  • The calculation gives us \( F_{net} = \sqrt{121378.89 + 223729} = \sqrt{345107.89} \approx 587.48 \ ext{N} \).
This approach allows us to see how individual force components contribute to the overall force acting on the object.

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Most popular questions from this chapter

A space traveler weighs 540.0 N on earth. What will the traveler weigh on another planet whose radius is twice that of earth and whose mass is three times that of earth?

The weight of an object is the same on two different planets. The mass of planet \(A\) is only sixty percent that of planet \(B .\) Find the ratio \(r_{A} / r_{B}\) of the radii of the planets.

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?

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) \(0 \mathrm{m} / \mathrm{s}^{2}.\)

A rocket of mass \(4.50 \times 10^{5} \mathrm{kg}\) is in flight. Its thrust is directed at an angle of \(55.0^{\circ}\) above the horizontal and has a magnitude of \(7.50 \times 10^{6} \mathrm{N} .\) Find the magnitude and direction of the rocket's acceleration. Give the direction as an angle above the horizontal.
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