Problem 29 - University Physics with Modern Physics

Question

A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 3.0 m in the direction of the initial velocity by a force with a magnitude of 10.0 N. (a) Use the work\(-\)energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

Step-by-Step Solution

Answer

Final speed is 4.95 m/s by both methods, matching results.

Step 1: Understand the Work-Energy Theorem

The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it is expressed as \( W = \Delta KE \), where \( W \) is the work done, and \( \Delta KE \) is the change in kinetic energy.

Step 2: Calculate Work Done

Work done \( W \) by a constant force is given by the equation \( W = F \cdot d \), where \( F \) is the force applied, and \( d \) is the distance over which the force is applied. Here, \( F = 10.0 \text{ N} \) and \( d = 3.0 \text{ m} \). So, \( W = 10.0 \times 3.0 = 30.0 \text{ J} \).

Step 3: Calculate Initial and Final Kinetic Energies

The initial kinetic energy \( KE_i \) is given by \( KE_i = \frac{1}{2} m v_i^2 \), where \( m = 7.00 \text{ kg} \) and \( v_i = 4.00 \text{ m/s} \). So, \( KE_i = \frac{1}{2} \times 7.00 \times 4.00^2 = 56.0 \text{ J} \). The final kinetic energy \( KE_f \) is \( KE_i + W = 56.0 + 30.0 = 86.0 \text{ J} \).

Step 4: Calculate Final Speed using Work-Energy Theorem

Use the formula for kinetic energy \( KE_f = \frac{1}{2} m v_f^2 \) to solve for the final speed \( v_f \). From \( KE_f = 86.0 \text{ J} \), we get \( 86.0 = \frac{1}{2} \times 7.00 \times v_f^2 \). Solving for \( v_f \), \( v_f = \sqrt{\frac{172.0}{7.00}} \approx 4.95 \text{ m/s} \).

Step 5: Calculate Acceleration Using Newton's Second Law

Newton's second law states \( F = m a \). Here, \( F = 10.0 \text{ N} \) and \( m = 7.00 \text{ kg} \). Therefore, the acceleration \( a = \frac{F}{m} = \frac{10.0}{7.00} \approx 1.43 \text{ m/s}^2 \).

Step 6: Use Kinematics to Find Final Speed

The kinematic equation \( v_f^2 = v_i^2 + 2 a d \) relates the initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), and distance \( d \). Substituting the known values, \( v_f^2 = 4.00^2 + 2 \times 1.43 \times 3.0 = 16.00 + 8.58 = 24.58 \). Solving for \( v_f \), \( v_f = \sqrt{24.58} \approx 4.95 \text{ m/s} \).

Step 7: Compare Results from Both Methods

Both the work-energy theorem and kinematics give the same final speed, \( v_f \approx 4.95 \text{ m/s} \). This confirms the consistency of the calculations using different approaches.

Key Concepts

Kinematics

Kinematics deals with the motion of objects. It's about describing how objects move, their speed, direction, and position over time. In this exercise, we're dealing with a wagon moving in a straight line on a horizontal surface. When exploring kinematics, we use equations to find unknown quantities like final speed or distance. These equations consider initial speed, acceleration, and distance traveled. For example, the equation \( v_f^2 = v_i^2 + 2ad \) allows us to find the final velocity \( v_f \) when the initial velocity \( v_i \), the acceleration \( a \), and the distance \( d \) are known.
The key is to apply these kinematic equations to make precise predictions about the motion of objects.

Newton's Second Law

Newton's second law is fundamental in physics, connecting the force applied to an object, its mass, and the resulting acceleration. It is formally stated as \( F = ma \), where \( F \) stands for force, \( m \) for mass, and \( a \) for acceleration. In the wagon example, with a force of 10.0 N and a mass of 7.00 kg, we use this law to determine the acceleration: \( a = \frac{F}{m} \). This leads to an acceleration of approximately 1.43 m/s².
Understanding Newton’s second law helps us predict how a given force will influence the motion of an object, which is essential for solving many physics problems.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, expressed mathematically as \( KE = \frac{1}{2}mv^2 \). In our exercise, kinetic energy helps us understand how the speed of the wagon is related to its energy. Initially, the wagon with a mass of 7.00 kg and speed of 4.00 m/s has a kinetic energy of 56.0 J. When work is done on the wagon by a force, this energy changes, demonstrating how kinetic energy is a crucial concept when evaluating how forces result in changes in speed.
Recognizing these energy transformations underlines the connection between force, motion, and the energy perspective in physics.

Acceleration

Acceleration is the rate at which an object's velocity changes over time. It's essential when we want to understand how quickly an object speeds up or slows down. In the context of this problem, we calculated acceleration using both Newton's second law and its effects in kinematic equations. We found that the acceleration generated by applying a 10 N force on the 7 kg wagon is 1.43 m/s². This acceleration tells us how the velocity of the wagon increases as it is moved along the frictionless surface.
Grasping acceleration is crucial for comprehending the stepping stone between force application and changes in an object's motion.

Frictionless Surface

A frictionless surface simplifies many physics problems, as it means there's no opposing force acting on a moving object due to friction. In this exercise, the lack of friction means the only horizontal force acting on the wagon is the applied force of 10 N. This setup allows us to apply the work-energy theorem and Newton's second law without needing to account for energy losses due to friction, making calculations straightforward.
Understanding the role of a frictionless surface helps us focus on core concepts like work, energy, and force, without the complexities introduced by friction.