Problem 22 - College Physics

Question

\(\bullet\) A 0.420 kg soccer ball is initially moving at 2.00 \(\mathrm{m} / \mathrm{s}\) . A soccer player kicks the ball, exerting a constant 40.0 \(\mathrm{N}\) force in the same direction as the ball's motion. Over what distance must her foot be in contact with the ball to increase the ball's speed to 6.00 \(\mathrm{m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

0.168 meters.

Step 1: Determine Initial and Final Kinetic Energy

Calculate the initial kinetic energy using the formula \( KE_i = \frac{1}{2} m v_i^2 \), where \( m = 0.420 \; \text{kg} \) and \( v_i = 2.00 \; \text{m/s} \). This gives \( KE_i = \frac{1}{2} \times 0.420 \times 2.00^2 = 0.840 \; \text{J} \). Next, calculate the final kinetic energy \( KE_f = \frac{1}{2} m v_f^2 \) with \( v_f = 6.00 \; \text{m/s} \), so \( KE_f = \frac{1}{2} \times 0.420 \times 6.00^2 = 7.56 \; \text{J} \).

Step 2: Calculate the Work Done

The work done on the ball is equal to the change in kinetic energy, \( W = KE_f - KE_i \). Substitute the values found to get \( W = 7.56 \; \text{J} - 0.840 \; \text{J} = 6.72 \; \text{J} \).

Step 3: Relate Work Done to Force and Distance

Use the work done formula \( W = F \cdot d \), where \( F = 40.0 \; \text{N} \). Solve for \( d \) (distance): \( d = \frac{W}{F} = \frac{6.72}{40.0} = 0.168 \; \text{m} \).

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that describes the energy an object possesses due to its motion. It is calculated using the formula:

• \( KE = \frac{1}{2} m v^2 \)

where \( m \) is the mass of the object in kilograms and \( v \) is the velocity of the object in meters per second.
The kinetic energy depends on both the mass and the square of the velocity, meaning small changes in speed can result in significant changes in kinetic energy. In our example, a soccer ball with a mass of 0.420 kg, initially moving at 2.00 m/s, has a kinetic energy of 0.840 J (joules).
When the ball's speed increases to 6.00 m/s, the kinetic energy increases to 7.56 J.
This change illustrates how the increase in velocity, even though only a factor of three, results in an increase of kinetic energy by a factor of nine, emphasizing the significance of the velocity component in the equation.

Force and Motion

Force is what causes an object to move or change its motion. In physics, when we talk about force and motion together, we're often discussing Newton's Laws of Motion. In this exercise, the soccer player exerts a force of 40.0 N on the ball, which is crucial for understanding how the ball's motion changes.
The application of force in the same direction as the motion allows the ball to accelerate, increasing its velocity from 2.00 m/s to 6.00 m/s.

• Force \( F \) is measured in newtons (N).
• The acceleration of the ball is due to this unbalanced force.

The player's kick provides a net force that overcomes any opposing forces, such as friction, helping the ball speed up. This process is a direct application of the work-energy theorem, where the work done by the force is converted into kinetic energy, resulting in increased speed.

Distance Calculation

To find out how far the soccer player's foot must be in contact with the ball, we use the concept of work, which in physics is the effort applied to move an object over a distance. This work is calculated by multiplying the force by the distance over which it is applied:

• Work \( W = F \times d \)

In this problem, we've determined that the work done (the change in kinetic energy) is 6.72 J, and the force applied is 40.0 N. To find the distance \( d \), we rearrange the formula to solve for \( d \): \\[ d = \frac{W}{F} \]Substitute the values to get:

• \( d = \frac{6.72 \, J}{40.0 \, N} = 0.168 \, m \)

This distance, 0.168 meters (or 16.8 centimeters), is the length over which the force must be applied to achieve the desired increase in speed. Understanding this calculation is key to visualizing how the application of constant force over a specific distance results in the increased kinetic energy of the soccer ball.