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Chapter 6: Problem 26

A soccer ball with mass \(0.420 \mathrm{~kg}\) is initially moving with speed \(2.00 \mathrm{~m} / \mathrm{s}\). A soccer player kicks the ball, exerting a constant force of magnitude \(40.0 \mathrm{~N}\) 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 \(6.00 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
The player's foot must be in contact with the ball over a distance of 0.168 m.

Step by step solution

01

Calculate Initial Kinetic Energy

First we need to calculate the initial kinetic energy of the ball. We can use the kinetic energy formula which is \(0.5*m*v^2\), where \(m\) is the mass and \(v\) is the speed. Substituting the values we get \(0.5*0.420*2^2 = 0.84 J\) (Joules).
02

Calculate Final Kinetic Energy

Next we calculate the final kinetic energy of the ball using the same formula but with the final speed (6.00 m/s). So we get \(0.5*0.420*6^2 = 7.56 J\).
03

Calculate Work Done

The work done by the force which is the change in kinetic energy is the difference between the final and initial kinetic energy. So \(\Delta K = 7.56 - 0.84 = 6.72 J\). The work done is equal to the force multiplied by the displacement, so \(W = F * d\). We rearrange the equation for displacement \(d = W / F\).
04

Find Displacement

Now we substitute the known values inside the equation: \(d = 6.72 / 40 = 0.168 m\). So the player's foot must be in contact with the ball over a distance of 0.168 m to increase the ball's speed as given in the problem.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It's an essential concept in physics that reveals how much work an object can do based on its mass and velocity. In mathematical terms, kinetic energy (KE) is expressed by the equation: \( KE = \frac{1}{2} mv^2 \), where \(m\) represents the mass of the object and \(v\) is its velocity.

In the context of the exercise, calculating the initial and final kinetic energies of the soccer ball helps us understand the energy change resulting from the player's kick. If the ball's speed increases, its kinetic energy also rises, signifying the ball can do more work moving forward. Recognizing these changes in kinetic energy is crucial for solving problems related to the work done on moving objects.
Force
Force is a vector quantity that describes the push or pull on an object, causing it to change its motion. It can cause an object to start moving, stop moving, or change its velocity. Forces are measured in Newtons (N) and mathematically expressed in Newton's second law as \( F = ma \) where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration.

In the exercise we discussed, the constant force exerted by the soccer player on the ball determines the amount of work done, which in turn dictates the increase in the ball's kinetic energy and velocity.
Displacement
Displacement refers to an object's overall change in position, moving from one point to another in a specific direction. It differs from distance as it is a vector quantity, which means it takes both magnitude and direction into account. In physics problems, displacement often underlines the 'path' taken by an object under the influence of a force.

The exercise in question details a situation where the soccer ball is moved through a specific displacement as a result of the force exerted by the player. This displacement is critical to calculate the work done by the force, impacting the ball's final velocity.
Work Done by Force
The 'work done' by a force is the energy transferred when a force acts upon an object to cause displacement. It is quantified as the product of force and displacement in the direction of the force and is measured in Joules (J). The equation for calculating work done is: \( W = F * d \) where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the displacement.

Understanding the Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.

In our exercise, we use this principle to determine the distance over which the player's foot must be in contact with the ball. By knowing the force applied and the change in kinetic energy, we can use the work-energy principle to find the required displacement to achieve the desired increase in the ball's speed.

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Most popular questions from this chapter

Your physics book is resting in front of you on a horizontal table in the campus library. You push the book over to your friend, who is seated at the other side of the table, 0.400 m north and 0.300 m east of you. If you push the book in a straight line to your friend, friction does \(-4.8 \mathrm{~J}\) of work on the book. If instead you push the book \(0.400 \mathrm{~m}\) due north and then \(0.300 \mathrm{~m}\) due east, how much work is done by friction?

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{~N}) \hat{\imath}-(40 \mathrm{~N}) \hat{\jmath}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{~m}) \hat{\imath}-(3.0 \mathrm{~m}) \hat{\jmath}\). How much work does the force you apply do on the grocery cart?

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{~m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{~N}+(3.0 \mathrm{~N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

A net horizontal force \(F\) is applied to a box with mass \(M\) that is on a horizontal, frictionless surface. The box is initially at rest and then moves in the direction of the force. After the box has moved a dis- tance \(D,\) the work that the constant force has done on it is \(W_{D}\) and the speed of the box is \(V\). The equation \(P=F v\) tells us that the instanta neous rate at which \(F\) is doing work on the box depends on the speed of the box. (a) At the point in the motion of the box where the force has done half the total work, and so has done work \(W_{D} / 2\) on the box that started from rest, in terms of \(V\) what is the speed of the box? Is the speed at this point less than, equal to, or greater than half the final speed? (b) When the box has reached half its final speed, so its speed is \(V / 2,\) how much work has been done on the box? Express your answer in terms of \(W_{D}\). Is the amount of work done to produce this speed less than, equal to, or greater than half the work \(W_{D}\) done for the full displacement \(D ?\)

CP You are pushing a large box across a frictionless floor by applying a constant horizontal force. If the box starts at rest, you have to do work \(W_{1}\) in order for the box to travel a distance \(d\) in time \(t .\) How much work would you have to do, in terms of \(W_{1}\), to make the box go the same distance in half the time?
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