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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 workenergy 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).

Short Answer

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

Step by step solution

01

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=ΔKE, where W is the work done, and ΔKE is the change in kinetic energy.
02

Calculate Work Done

Work done W by a constant force is given by the equation W=Fd, where F is the force applied, and d is the distance over which the force is applied. Here, F=10.0 N and d=3.0 m. So, W=10.0×3.0=30.0 J.
03

Calculate Initial and Final Kinetic Energies

The initial kinetic energy KEi is given by KEi=12mvi2, where m=7.00 kg and vi=4.00 m/s. So, KEi=12×7.00×4.002=56.0 J. The final kinetic energy KEf is KEi+W=56.0+30.0=86.0 J.
04

Calculate Final Speed using Work-Energy Theorem

Use the formula for kinetic energy KEf=12mvf2 to solve for the final speed vf. From KEf=86.0 J, we get 86.0=12×7.00×vf2. Solving for vf, vf=172.07.004.95 m/s.
05

Calculate Acceleration Using Newton's Second Law

Newton's second law states F=ma. Here, F=10.0 N and m=7.00 kg. Therefore, the acceleration a=Fm=10.07.001.43 m/s2.
06

Use Kinematics to Find Final Speed

The kinematic equation vf2=vi2+2ad relates the initial velocity vi, final velocity vf, acceleration a, and distance d. Substituting the known values, vf2=4.002+2×1.43×3.0=16.00+8.58=24.58. Solving for vf, vf=24.584.95 m/s.
07

Compare Results from Both Methods

Both the work-energy theorem and kinematics give the same final speed, vf4.95 m/s. This confirms the consistency of the calculations using different approaches.

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Key Concepts

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

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 vf2=vi2+2ad allows us to find the final velocity vf when the initial velocity vi, 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=Fm. 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=12mv2. 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.

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Most popular questions from this chapter

A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

You are applying a constant horizontal force F=(8.00N)ı^+(3.00N)ȷ^ to a crate that is sliding on a factory floor. At the instant that the velocity of the crate is υ=(3.20m/s)ı^+(2.20m/s)ȷ^, what is the instantaneous power supplied by this force?

A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

Use the workenergy theorem to solve each of these problems. You can use Newton's laws to check your answers. (a) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping? (b) Suppose the rough patch in part (a) was only 2.90 m long. How fast would the skier be moving when she reached the end of the patch? (c) At the base of a frictionless icy hill that rises at 25.0 above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?

An object has several forces acting on it. One of these forces is F=axyı^, a force in the x-direction whose magnitude depends on the position of the object, with α=2.50N/m2. Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point (x=0, y=3.00 m) and moves parallel to the x-axis to the point (x=2.00 m, y=3.00 m). (b) The object starts at the point (x=2.00 m, y=0) and moves in the y-direction to the point (x=2.00 m, y=3.00 m). (c) The object starts at the origin and moves on the line y=1.5x to the point (x=2.00 m, y=3.00 m).

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