Problem 25 - University Physics with Modern Physics

Question

A 12 -pack of Omni-Cola (mass 4.30 \(\mathrm{kg}\) ) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 \(\mathrm{m}\) by a trained dog that exerts a horizontal force with magnitude 36.0 \(\mathrm{N}\) . 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) 4.46 m/s, (b) 3.62 m/s.

Step 1: Understanding the Problem

We need to find the final speed of a 12-pack pushed by a dog for 1.20 m. In part (a), there is no friction, while in part (b), there is kinetic friction with a coefficient of 0.30. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy.

Step 2: Calculating Work Done by the Dog (No Friction)

First, let's calculate the work done by the dog. Work done (W) is given by the formula \( W = F \times d \times \cos(\theta) \), where \( F = 36.0 \ \text{N} \), \( d = 1.20 \ \text{m} \), and \( \theta = 0^\circ \) because the force is horizontal. Thus, \( W = 36.0 \times 1.20 \times \cos(0) = 36.0 \times 1.20 = 43.2 \, \text{J} \).

Step 3: Applying the Work-Energy Theorem (No Friction)

According to the work-energy theorem, the work done is equal to the change in kinetic energy: \( W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \). Since the 12-pack starts from rest, \( u = 0 \). Therefore, \( 43.2 = \frac{1}{2} \times 4.30 \times v^2 \). Solving for \( v \), we get \( v = \sqrt{\frac{2 \times 43.2}{4.30}} \), which results in \( v \approx 4.46 \, \text{m/s} \).

Step 4: Calculating Frictional Force (With Friction)

Now, for part (b), we consider friction. The frictional force (\( f_k \)) is calculated using \( f_k = \mu_k \times m \times g \), where \( \mu_k = 0.30 \), \( m = 4.30 \ \text{kg} \), and \( g = 9.8 \ \text{m/s}^2 \). Thus, \( f_k = 0.30 \times 4.30 \times 9.8 = 12.636 \, \text{N} \).

Step 5: Calculating Net Work Done (With Friction)

The net work done is the work by the dog minus the work done against friction. Work against friction is \( f_k \times d = 12.636 \times 1.20 = 15.1632 \, \text{J} \). The net work done is \( 43.2 - 15.1632 = 28.0368 \, \text{J} \).

Step 6: Applying the Work-Energy Theorem (With Friction)

Using the net work done and the work-energy theorem, \( 28.0368 = \frac{1}{2} \times 4.30 \times v^2 \). Solving for \( v \), we have \( v = \sqrt{\frac{2 \times 28.0368}{4.30}} \), which results in \( v \approx 3.62 \, \text{m/s} \).

Key Concepts

Kinetic Friction

When objects slide against each other, they encounter resistance known as kinetic friction. This force opposes the motion of the object and acts parallel to the contact surface. In the case of moving a 12-pack of Omni-Cola across a floor with friction, we account for this resistance using the coefficient of kinetic friction \( \mu_k \). This dimensionless value represents how grippy two surfaces are when sliding past each other. The higher the \( \mu_k \), the more frictional force there is. Here, with \( \mu_k \) given as 0.30, friction significantly affects motion, slowing our 12-pack down.

Work Done

Work is an essential part of mechanics representing how much energy is transferred through force acting over a distance. Calculating work involves the formula \( W = F \times d \times \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved by the object, and \( \theta \) is the angle between the force and the direction of motion. For our exercise, since the force exerted by the dog is horizontal, \( \theta = 0^\circ \), making \( \cos(0^\circ) = 1 \). This simplifies our calculation to \( W = F \times d \). This quick computation assists in determining how much energy the dog transfers to the Omni-Cola packet, paving the way for determining kinetic energy changes.

Kinetic Energy

Kinetic energy is the energy of an object due to its motion, calculated using the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) represents mass and \( v \) stands for velocity. As work is done on the Omni-Cola, its kinetic energy changes. Initially, it starts from rest, so its initial kinetic energy is zero. All work done goes into raising its kinetic energy. Thus, the calculation of kinetic energy helps us find out how fast the 12-pack moves after being pushed by the dog, given the conditions of friction or the lack thereof. Understanding kinetic energy variations explains the effects of net forces on the object's speed.

Net Work

Net work determines the total work done on an object after considering all forces, including friction. In this context, the net work is the initial work done by the dog minus the work associated with overcoming friction. Here's how it works:
- Calculate the work done by friction: \( f_k \times d \).
- Subtract this from the total work done by the dog to find the net work.
This net work practically determines the remaining energy converting into kinetic energy, influencing the speed of the object. It elucidates how various forces influence the system, shedding light on the power distribution and outcome of different friction scenarios.

Force Calculation

Calculating forces correctly is fundamental to understanding motion. The dog applies a force that drives the system, initiating the slide of the 12-pack. Use \( F = m \times a \), where \( F \) is the force, \( m \) is mass, and \( a \) is acceleration, to determine how much force is necessary to accelerate the Omni-Cola pack. In our situation, knowing the force exerted (36.0 N) is crucial as it accounts for the primary factor driving the system. In combination with frictional force, it dictates how effectively the 12-pack gains speed over the given distance. Mastery of force calculation reveals the required energy input and resistance encountered, mapping how different elements interplay in motion.