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Chapter 6: Problem 27

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 m by a trained dog that exerts a horizontal force with magnitude 36.0 N. Use the work\(-\)energy theorem to find the final speed of the 12-pack if (a) there is no friction between the 12-pack and the floor, and (b) the coefficient of kinetic friction between the 12-pack and the floor is 0.30.

Short Answer

Expert verified
(a) 4.48 m/s, (b) 3.61 m/s.

Step by step solution

01

Understanding the Work-Energy Theorem

The work-energy theorem states that the work done by all external forces on an object equals the change in the object's kinetic energy. Mathematically, it's given by \( W = \Delta KE = \frac{1}{2} m v^2_f - \frac{1}{2} m v^2_i \), where \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.
02

Calculate Work Done when No Friction

When there is no friction, the only work done is by the dog. This is given by \( W = F \cdot d \), where \( F = 36.0 \text{ N} \) and \( d = 1.20 \text{ m} \). Thus, \( W = 36.0 \times 1.20 = 43.2 \text{ J} \).
03

Apply Work-Energy Theorem with No Friction

Apply the work-energy theorem: \( 43.2 = \frac{1}{2} \times 4.30 \times v_f^2 - 0 \), since initial velocity \( v_i = 0 \) (at rest). Solve for \( v_f \): \[ 43.2 = \frac{1}{2} \times 4.30 \times v_f^2 \] and \[ v_f = \sqrt{\frac{43.2 \times 2}{4.30}} \approx 4.48 \text{ m/s}. \]
04

Calculate Frictional Force for Friction Scenario

The frictional force \( f_k \) is given by \( f_k = \mu_k \times N \), where \( \mu_k = 0.30 \) and \( N = m \cdot g = 4.30 \times 9.8 \text{ N} \approx 42.14 \text{ N} \). Thus, \( f_k = 0.30 \times 42.14 \approx 12.64 \text{ N} \).
05

Calculate Net Work Done with Friction

The net work done \( W_{net} \) is \( W_{net} = F \cdot d - f_k \cdot d = 43.2 - 12.64 \times 1.20 \). Thus, \( W_{net} = 43.2 - 15.168 = 28.032 \text{ J} \).
06

Apply Work-Energy Theorem with Friction

Using the work-energy theorem: \( 28.032 = \frac{1}{2} \times 4.30 \times v_f^2 \). Solve for \( v_f \): \[ 28.032 = \frac{1}{2} \times 4.30 \times v_f^2 \] and \[ v_f = \sqrt{\frac{28.032 \times 2}{4.30}} \approx 3.61 \text{ m/s}. \]

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is one of the pivotal concepts in physics, defining how much work an object can produce based on its speed and mass. The equation to calculate kinetic energy (KE) is given by:\[KE = \frac{1}{2} m v^2\]- \( m \) is the mass of the object- \( v \) is the velocity of the object
For example, in the exercise with the Omni-Cola, initially the pack is at rest, meaning its kinetic energy is zero. As the dog pushes it, the pack gains velocity, thus increasing its kinetic energy. The increase in kinetic energy directly relates to the work done on the object, which is why the work-energy theorem plays a crucial role in calculating the final speed.
The Role of Friction
Friction is a force that opposes the motion of objects. It's essential in various real-world scenarios, as it affects how objects move across surfaces. When solving physics problems, understanding friction is crucial to accurately determine the net work done on a moving object. In the given exercise, friction is considered under scenario (b), where it acts as a counteracting force against the dog's push.
The force of kinetic friction ( f_k ) is calculated using:\[f_k = \mu_k \times N\]- \( \mu_k \) is the coefficient of kinetic friction- \( N \) is the normal force, which is the weight of the object when it's on a flat surface
By calculating the frictional force, one can determine how much of the applied force is used to overcome this friction, influencing the final speed of the Coca-Cola pack. Understanding how friction changes the net work done is pivotal for engineering and safety reasons in everyday applications.
Navigating Physics Problems with Work-Energy Theorem
Physics problems often require an understanding of how different forces and energies interact. The work-energy theorem is a powerful tool that connects the dots between force, motion, and energy in a problem. It expresses that the work done on an object results in a change of its kinetic energy, highlighting a direct relation between work performed and the motion achieved. Let鈥檚 summarize its application:
  • Identify the forces acting on the object.
  • Calculate the work done by each force to understand the total work.
  • Use the work-energy theorem to find the change in kinetic energy.

For the given Omni-Cola example, calculating without and with friction showcases how external forces like friction impact the result. By mastering this approach, students can tackle complex physics problems efficiently, predicting real-world outcomes with ease.

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Most popular questions from this chapter

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 W. If she expends a total of 1.1 \(\times\) 10\(^7\) J of energy in a 24-hour day, how much of the day did she spend walking?

A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_k\) \(=\) 0.30. Use the work\(-\)energy theorem to find how far the textbook moves from its initial position before it comes to rest.

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?
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