Work And Energy 4.b Choosing Systems

Let's delve into the crucial aspect of system selection when analyzing work and energy scenarios, a cornerstone of physics that bridges the gap between abstract theory and real-world applications. Understanding how to effectively define a system is paramount to correctly applying the work-energy theorem and accurately predicting the behavior of physical phenomena.

Work and Energy: The Art of Choosing Systems

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This powerful tool simplifies the analysis of many physical situations, but its correct application hinges on the careful selection of the "system." The system is the specific object or collection of objects whose motion and energy changes we are interested in. Everything else is considered the "environment." Choosing the right system is not just a matter of convenience; it is a critical step that dictates which forces do work on the system, which forces are internal, and ultimately, how easily the problem can be solved.

Why Does System Choice Matter?

The choice of a system dictates:

• Which forces are considered external and do work: External forces are those exerted on the system by the environment. These are the forces that can change the system's kinetic energy. Internal forces, on the other hand, are forces exerted between objects within the system. While internal forces can do work, they do not contribute to the net work done on the system and therefore do not directly change the system's total kinetic energy (though they can redistribute energy within the system).
• Whether energy is conserved: If the only forces doing work are conservative forces (like gravity or the force from an ideal spring), the total mechanical energy (kinetic plus potential) of the isolated system is conserved. A isolated system is one where no external forces do work. If external forces do work, energy is not conserved within the system, but the work-energy theorem still applies.
• The complexity of the problem: A well-chosen system can dramatically simplify the analysis, reducing the number of forces to consider and making the application of the work-energy theorem more straightforward. A poorly chosen system can lead to unnecessary complexity and even incorrect results.

Guidelines for Choosing a System

Here are some guidelines to help you choose the most appropriate system for a given problem:

1. Identify the Object(s) of Interest: Start by clearly defining what you want to study. Are you interested in the motion of a single block, a car, a roller coaster, or a more complex assembly of objects? This is the foundation for defining your system.
2. Consider the Forces Involved: Identify all the forces acting on the object(s) of interest. Distinguish between forces exerted by objects within the potential system and forces exerted by the environment on the potential system. This is crucial for determining which forces will do work on the system.
3. Think About Conservation of Energy: Ask yourself: Is mechanical energy conserved in the situation? If only conservative forces are doing work, consider including all relevant objects within your system to make it an isolated system where energy is conserved. If non-conservative forces are present, you may need to define a smaller system and account for the work done by those forces.
4. Simplify Where Possible: Aim for the simplest system that allows you to answer the question at hand. Sometimes, including unnecessary objects in the system can complicate the analysis without providing any additional insight.
5. Be Consistent: Once you've defined your system, stick with it throughout the entire analysis. Avoid changing the system mid-problem, as this can lead to inconsistencies and errors.

Examples Illustrating System Selection

Let's illustrate the importance of system selection with several examples:

Example 1: A Block Sliding Down an Inclined Plane

A block of mass m slides down a frictionless inclined plane of height h. Determine the speed of the block at the bottom of the plane.

• System 1: The Block Alone
  • Environment: The Earth (providing gravity) and the inclined plane (providing the normal force).
  • Forces doing work: Gravity (conservative force). The normal force does no work because it's perpendicular to the displacement.
  • Analysis: The work done by gravity is mgh. By the work-energy theorem: mgh = (1/2)mv² - 0. Therefore, v = √(2gh).
• System 2: The Block and the Earth
  • Environment: The inclined plane (providing the normal force).
  • Forces doing work: Only the normal force (does no work). Gravity is now an internal force within the system.
  • Analysis: Since only the normal force does work (and it does zero work), external work is zero. We can use conservation of mechanical energy: The initial potential energy of the block-Earth system (relative to the bottom of the incline) is mgh. The final kinetic energy of the block is (1/2)mv². Conserving energy: mgh = (1/2)mv². Therefore, v = √(2gh).

In this case, both systems lead to the correct answer. However, System 2 allows us to use conservation of energy directly, which is often simpler than calculating the work done by gravity.

Example 2: A Block Pulled Across a Rough Surface

A block of mass m is pulled across a rough horizontal surface by a force F at an angle θ above the horizontal. The coefficient of kinetic friction between the block and the surface is μk. Determine the work done by each force acting on the block as it moves a distance d.

• System: The Block
  • Environment: The Earth (providing gravity), the pulling force F, and the surface (providing the normal force and friction).
  • Forces doing work:
    • Pulling force F: Work done is Fdcos(θ).
    • Friction force fk: Work done is -fkd = -μk N d, where N is the normal force.
    • Gravity: No work done (perpendicular to the displacement).
    • Normal force: No work done (perpendicular to the displacement).
  • Analysis: To find the normal force, we sum forces in the vertical direction: N + Fsin(θ) - mg = 0. Therefore, N = mg - Fsin(θ). The work done by friction is -μk(mg - Fsin(θ))d. The net work done on the block is Fdcos(θ) - μk(mg - Fsin(θ))d.

In this case, defining the system as only the block is the most straightforward approach. If we tried to include the surface in the system, we would have to account for the internal work done within the surface due to friction, which is significantly more complex.

Example 3: A Collision Between Two Blocks

Two blocks, A and B, with masses mA and mB, respectively, collide on a frictionless horizontal surface. Block A has an initial velocity vA and block B is initially at rest. After the collision, the blocks stick together and move as a single unit. Determine the final velocity of the combined blocks.

• System 1: Block A Alone
  • Environment: Block B (providing the collision force).
  • Forces doing work: The collision force. This force is difficult to determine precisely.
  • Analysis: Applying the work-energy theorem to block A alone is not helpful because we don't know the collision force or the distance over which it acts.
• System 2: Block B Alone
  • Environment: Block A (providing the collision force).
  • Forces doing work: The collision force. Again, this force is difficult to determine.
  • Analysis: Similar to System 1, applying the work-energy theorem to block B alone is not helpful.
• System 3: Block A and Block B Together
  • Environment: None (assuming a frictionless surface).
  • Forces doing work: No external forces do work. The collision force is now an internal force within the system.
  • Analysis: Since no external forces do work, momentum is conserved. The initial momentum of the system is mA vA. The final momentum of the system is (mA + mB) vf, where vf is the final velocity of the combined blocks. Conserving momentum: mA vA = (mA + mB) vf. Therefore, vf = (mA vA) / (mA + mB).

In this scenario, choosing the combined blocks as the system is crucial. This allows us to treat the collision force as an internal force, meaning no external work is done, and we can apply the conservation of momentum. Attempting to analyze each block separately would be much more difficult.

Example 4: A Mass-Spring System (Vertical)

Consider a mass m attached to a spring with spring constant k. The mass is released from rest at the spring's unstretched length. Determine the maximum extension of the spring.

• System 1: The Mass Alone
  • Environment: The Earth (gravity) and the spring (spring force).
  • Forces doing work: Gravity (conservative) and spring force (conservative).
  • Analysis: We could use the work-energy theorem, calculating the work done by gravity and the spring force. However, this requires careful consideration of the displacement and the changing spring force.
• System 2: The Mass and the Spring
  • Environment: The Earth (gravity).
  • Forces doing work: Gravity on the mass. Spring force is now internal.
  • Analysis: We can use conservation of energy. Let x be the maximum extension of the spring.
    • Initial state: Mass at rest, spring unstretched. Gravitational potential energy (relative to the initial position): 0. Elastic potential energy: 0. Kinetic energy: 0.
    • Final state: Mass momentarily at rest at maximum extension. Gravitational potential energy: -mgx. Elastic potential energy: (1/2)kx². Kinetic energy: 0.
    • Conserving energy: 0 = -mgx + (1/2)kx². Solving for x: x = 2mg/k.
• System 3: The Mass, the Spring, and the Earth
  • Environment: None. The whole universe except the mass, spring, and Earth is the environment! (This isn't helpful)
  • Forces doing work: No external forces do work. Gravity and spring forces are both internal.
  • Analysis: We can still use conservation of energy, but now the gravitational potential energy is also internal to the system. We define the gravitational potential energy relative to a reference point (say, the initial position of the mass). The change in gravitational potential energy is still -mgx, and the change in elastic potential energy is (1/2)kx², leading to the same result as System 2.

In this example, System 2 is often the most convenient because it clearly separates the work done by gravity from the internal energy stored in the spring. However, System 3 works perfectly well if you're comfortable defining the gravitational potential energy as part of the system's internal energy.

Example 5: A Car Braking

A car of mass m is traveling at an initial speed v0 and brakes to a stop over a distance d. Determine the average braking force.

• System: The Car
  • Environment: The road (providing the friction/braking force) and the Earth (gravity).
  • Forces doing work: The braking force (friction). Gravity and the normal force do no work.
  • Analysis: The initial kinetic energy of the car is (1/2)mv0². The final kinetic energy is 0. The work done by the braking force Fb is -Fbd. By the work-energy theorem: -Fbd = 0 - (1/2)mv0². Therefore, Fb = (1/2)mv0² / d.

Choosing only the car as the system allows for a direct application of the work-energy theorem. Including the road would complicate the analysis without providing any additional information.

Common Pitfalls to Avoid

• Forgetting Forces: Ensure that you identify all external forces acting on your chosen system. Missing a force will lead to an incorrect calculation of the net work done.
• Confusing Internal and External Forces: Clearly distinguish between forces acting within the system and forces exerted by the environment. Only external forces can change the total kinetic energy of the system.
• Changing the System Mid-Problem: Define your system at the beginning of the problem and stick with it. Changing the system in the middle of the analysis can lead to inconsistencies and errors.
• Incorrectly Applying Conservation of Energy: Remember that energy is only conserved in isolated systems where no external forces do work. If non-conservative forces are present (like friction), you must account for the work done by these forces.
• Assuming Forces Do Work When They Don't: A force only does work if it has a component parallel to the displacement of the object. Forces perpendicular to the displacement, like the normal force on a horizontally moving object, do no work.

Advanced Considerations

• Variable Forces: If the forces doing work are not constant, you may need to use integration to calculate the work done. The work-energy theorem still applies, but the calculation of the work may be more involved.
• Rotational Motion: When dealing with rotating objects, you need to consider rotational kinetic energy and the work done by torques. The system selection principles remain the same, but the analysis becomes more complex.
• Thermodynamics: In thermodynamic systems, energy can be exchanged in the form of heat and work. The first law of thermodynamics is a statement of energy conservation that includes these forms of energy transfer.

Conclusion

Choosing the right system is a fundamental skill in applying the work-energy theorem effectively. By carefully considering the objects of interest, the forces involved, and the potential for energy conservation, you can simplify complex problems and gain a deeper understanding of the underlying physics. Remember to be consistent, avoid common pitfalls, and practice applying these principles to a variety of scenarios. The ability to skillfully define a system is the key to unlocking the power of work and energy principles in solving a wide range of physics problems. Mastering this skill will not only improve your problem-solving abilities but also deepen your conceptual understanding of energy and its role in the physical world.