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Chapter 6: Problem 27

A little red wagon with mass 7.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 \(\mathrm{m} / \mathrm{s}\) and then is pushed 3.0 \(\mathrm{m}\) in the direction of the initial velocity by a force with a magnitude of 10.0 \(\mathrm{N}\) . (a) Use the work-energy theorem to calculare the wagon's final speed. (b) Cal- culate 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
The wagon's final speed is 4.95 m/s using both methods.

Step by step solution

01

Identify Given Values

First, gather all the relevant data from the problem. The mass of the wagon is 7.00 \(\mathrm{kg}\), the initial speed \(v_i\) is 4.00 \(\mathrm{m/s}\), the displacement \(s\) is 3.0 \(\mathrm{m}\), and the force \(F\) applied is 10.0 \(\mathrm{N}\).
02

Apply the Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy. Calculate the work done by the force: \[ W = F \times s = 10.0 \mathrm{\, N} \times 3.0 \mathrm{\, m} = 30.0 \mathrm{\, J} \]
03

Calculate the Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) is equal to the work done on the wagon. Thus, \( \Delta KE = W = 30.0 \mathrm{\, J} \). The initial kinetic energy \( KE_i = \frac{1}{2}mv_i^2 = 0.5 \times 7.00 \times 4.00^2 \).
04

Find Initial and Final Kinetic Energies

Calculate the initial kinetic energy: \[ KE_i = 0.5 \times 7.00 \times 4.00^2 = 56.0 \mathrm{\, J} \] Use \( \Delta KE = KE_f - KE_i = 30.0 \mathrm{\, J} \) to find the final kinetic energy: \( KE_f = KE_i + \Delta KE = 56.0 + 30.0 = 86.0 \mathrm{\, J} \).
05

Calculate the Final Velocity (Part a)

Use the final kinetic energy to find the final speed \(v_f\). \[ KE_f = \frac{1}{2}mv_f^2 = 86.0 \mathrm{\, J} \rightarrow \frac{1}{2}\times7.00\times v_f^2 = 86.0 \] Solve for \(v_f\): \[ v_f = \sqrt{\frac{2\times86.0}{7.00}} = 4.95 \mathrm{\, m/s} \].
06

Calculate Acceleration (Part b)

Using Newton's second law \( F = m\times a \) to find acceleration \(a\): \[ a = \frac{F}{m} = \frac{10.0}{7.0} = 1.43 \mathrm{\, m/s^2} \].
07

Use Kinematic Equations to Find Final Velocity

With acceleration known, use the kinematic equation \(v_f^2 = v_i^2 + 2as\) to find \(v_f\). \[ v_f^2 = 4.00^2 + 2\times1.43\times3.0 \] \[ v_f = \sqrt{16 + 8.58} = 4.95 \mathrm{\, m/s} \].
08

Compare Results

The final velocities from both methods are 4.95 \(\mathrm{m/s}\). This shows consistency between the work-energy method and the kinematic approach.

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Key Concepts

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

Work-Energy Theorem
The Work-Energy Theorem is a powerful principle in physics that connects the work done on an object to its kinetic energy. This theorem states that the work done by all external forces acting on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as:
  • \[ W = \Delta KE \]
In the context of our red wagon problem, the work done by the 10.0 N force over a displacement of 3.0 m is calculated as 30.0 J. Since the surface is frictionless, this work directly translates into changing the kinetic energy of the wagon.

The initial kinetic energy (\( KE_i \)) can be determined using the formula \( KE = \frac{1}{2}mv^2 \). For the initial speed (4.0 m/s) and mass (7.0 kg), the initial kinetic energy is 56.0 J. The work done (30.0 J) is added to this initial kinetic energy to find the final kinetic energy (86.0 J). From here, we can re-arrange the kinetic energy formula to solve for the final speed, confirming it as 4.95 m/s.
Kinematic Equations
Kinematic equations are equations that describe the motion of objects using variables such as velocity, acceleration, displacement, and time. One of the most commonly used kinematic equations is:
  • \[ v_f^2 = v_i^2 + 2as \]
In the problem at hand, once the acceleration is known, this equation becomes key in confirming the final speed calculated via the work-energy theorem.

The initial speed \( v_i \) is 4.0 m/s, and the acceleration \( a \) is determined to be 1.43 m/s虏. Plugging in the values, and the displacement \( s = 3.0 \) m, the kinematic equation also gives a final velocity \( v_f \) of 4.95 m/s.

This confirms the consistency of the results from the two different approaches in analyzing the same physical situation.
Newton's Second Law
Newton's Second Law of Motion is fundamental to solving many physics problems. It tells us that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
  • \[ F = ma \]
For the red wagon, this law is used to calculate the acceleration produced by the applied force. With a force of 10.0 N and a mass of 7.0 kg, the acceleration can be found using the rearranged form of Newton's Second Law:
  • \[ a = \frac{F}{m} = \frac{10.0}{7.0} \approx 1.43 \mathrm{\, m/s^2} \]
This calculated acceleration is critical in using the kinematic equations to determine the wagon's final speed.

By understanding Newton's Second Law, you can appreciate how forces influence the motion of objects, contributing to the cornerstone of dynamic analysis in physics.

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Most popular questions from this chapter

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k=200 \mathrm{N} / \mathrm{m}\) and is compressed 0.025 \(\mathrm{m}\) . The spring is released and accelerates the block along a horizontal surface. You can 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 1.50 -kg book is sliding along a rough horizontal surface. At point \(A\) it is moving at \(3.21 \mathrm{m} / \mathrm{s},\) and at point \(B\) it has slowed to 1.25 \(\mathrm{m} / \mathrm{s}\) (a) How much work was done on the book between \(A\) and \(B ?\) (b) If \(-0.750 \mathrm{J}\) of work is done on the book from \(B\) to \(C\) , how fast is it moving at point \(C ?\) (c) How fast would it be moving at \(C\) if \(+0.750 \mathrm{J}\) of work were done on it from \(B\) to \(C\) ?

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring with force constant \(k=40.0 \mathrm{N} / \mathrm{cm}\) and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 \(\mathrm{kg}\) are pushed against the other end, compressing the spring 0.375 \(\mathrm{m}\) . The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 \(\mathrm{m} ?\)

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{N}\) , one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker 0.75 \(\mathrm{km}\) toward the north. What is the total work they do on the supertanker?

BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.
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