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A soccer ball with mass 0.420 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?

Short Answer

Expert verified
The player's foot must be in contact with the ball for 0.168 meters.

Step by step solution

01

Apply the work-energy principle

According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. This can be expressed as: \( W = \Delta KE \). In this situation, \( W \) is the work done by the force of 40.0 N, and \( \Delta KE \) is the change in kinetic energy of the ball.
02

Calculate initial and final kinetic energies

The initial kinetic energy \( KE_i \) of the ball can be calculated using the formula \( KE_i = \frac{1}{2}mv_i^2 \), where \( m = 0.420 \text{ kg} \) and \( v_i = 2.00 \text{ m/s} \). So, \( KE_i = \frac{1}{2} \times 0.420 \times (2.00)^2 = 0.84 \text{ J} \).The final kinetic energy \( KE_f \) is calculated with \( v_f = 6.00 \text{ m/s} \): \( KE_f = \frac{1}{2} \times 0.420 \times (6.00)^2 = 7.56 \text{ J} \).
03

Find change in kinetic energy

The change in kinetic energy \( \Delta KE \) is \( KE_f - KE_i \). Substituting the known values: \( \Delta KE = 7.56 \text{ J} - 0.84 \text{ J} = 6.72 \text{ J} \).
04

Use work done formula

The work done \( W \) by the force is equal to the product of the force and the distance over which it acts. This is expressed as \( W = F \times d \). Given \( F = 40.0 \text{ N} \) and \( W = 6.72 \text{ J} \), we can solve for \( d \) as \( d = \frac{W}{F} = \frac{6.72}{40.0} \).
05

Calculate the distance

By dividing the work done by the force, \( d = \frac{6.72}{40.0} = 0.168 \text{ m} \). The player's foot must remain in contact with the ball for a distance of 0.168 meters to achieve this speed increase.

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Key Concepts

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

Kinetic Energy
Kinetic energy describes the energy that an object possesses because of its motion. It is a fundamental concept in physics, as it determines the capacity for doing work based on an object's mass and speed. The formula to calculate kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the speed of the object.
In the case of our soccer ball problem, we find the initial kinetic energy using its mass \( 0.420 \text{ kg} \) and initial speed \( 2.00 \text{ m/s} \). After the player's kick, the final kinetic energy is computed with the ball's new speed \( 6.00 \text{ m/s} \). Understanding these calculations is essential for applying the work-energy principle effectively.
Work Done
The concept of work done ties directly into energy transfer. In physics, work is accomplished when a force causes an object to move. Calculated as:\[ W = F \cdot d \]where \( F \) is the force applied, and \( d \) is the distance over which the force acts.
In our exercise, the work performed by the player on the soccer ball results in a change in kinetic energy. The player's force of 40.0 N increases the ball's speed via work done over a particular distance. By solving for this distance using the relationship between work and energy change, we understand that the player's kick transferred energy from the foot to the ball, evidenced by increased speed.
Force and Motion
Force is the interaction that alters an object's motion. According to Newton's second law of motion, force is connected to how an object moves or accelerates, described by:\[ F = m \cdot a \]where \( a \) is the acceleration. While this equation isn't directly used in our problem, it's crucial to understand this relationship.
The application of force by the soccer player on the ball changes the ball's velocity. The constant force of 40.0 N, applied in the direction of motion, shows how force influences motion, leading to an acceleration that increases the ball's speed from 2.00 m/s to 6.00 m/s. Understanding force allows us to see how interactions generate motion or change in speed.
Physics Problem Solving
Problem-solving in physics requires a systematic approach. Let's break down how the work-energy principle helps solve the soccer ball problem:
  • Identify the given values: mass, initial and final velocities, and force.
  • Compute initial and final kinetic energies to understand energy changes.
  • Use relationships known from the work-energy principle to relate these energy changes to the work done.
  • Set up equations with known values to solve for the unknown, in this case, the distance.
This structured approach, using physics principles, aids in solving problems by organizing information and applying relevant formulas. By breaking down the steps, students learn to approach complex scenarios with clarity and confidence.

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Most popular questions from this chapter

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40.0\(^\circ\) above the horizontal. The glider has mass 0.0900 kg. The spring has \(k\) = 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed to 1.65 m/s due to a friction force that is 25% of her weight. Use the work\(-\)energy theorem to find the length of this rough patch.

A proton with mass 1.67 \(\times\) 10\(^{-27}\) kg is propelled at an initial speed of 3.00 \(\times\) 10\(^5\) m/s directly toward a uranium nucleus 5.00 m away. The proton is repelled by the uranium nucleus with a force of magnitude \(F = \alpha/x^2\), where \(x\) is the separation between the two objects and \(\alpha = 2.12 \times 10^{-26} \, \mathrm{N} \cdot \mathrm{m}^2\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10}\) m from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again 5.00 m away from the uranium nucleus?

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k\) = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

A 6.0-kg box moving at 3.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. Use the work\(-\)energy theorem to find the maximum compression of the spring.

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