Problem 27 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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, \( W = 36.0 \times 1.20 = 43.2 \text{ J} \).

Step 3: 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 \[ v_f = \sqrt{\frac{43.2 \times 2}{4.30}} \approx 4.48 \text{ m/s}. \]

Step 4: 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, \( f_k = 0.30 \times 42.14 \approx 12.64 \text{ N} \).

Step 5: 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, \( W_{net} = 43.2 - 15.168 = 28.032 \text{ J} \).

Step 6: 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 \[ v_f = \sqrt{\frac{28.032 \times 2}{4.30}} \approx 3.61 \text{ m/s}. \]

Key Concepts

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’s 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.