Work And Energy Diagram Skills Answers

Mastering Work and Energy Diagrams: A Comprehensive Guide with Solutions

Work and energy diagrams are powerful tools in physics that allow us to visualize and analyze the energy transformations occurring within a system. They provide a clear and concise way to track energy input, output, and storage, making it easier to solve complex physics problems. This comprehensive guide will delve into the fundamentals of work and energy diagrams, illustrate their construction, and provide step-by-step solutions to various example problems.

Introduction to Work and Energy Concepts

Before we dive into the diagrams themselves, let's solidify our understanding of the underlying concepts:

• Work (W): Work is done when a force causes displacement. Mathematically, it is defined as the product of the force component along the direction of displacement and the magnitude of the displacement. W = Fd cos θ, where F is the force, d is the displacement, and θ is the angle between the force and displacement vectors. Work is a scalar quantity, measured in Joules (J).
• Energy (E): Energy is the capacity to do work. It exists in various forms, including:
  • Kinetic Energy (KE): Energy possessed by an object due to its motion. KE = 1/2 mv², where m is the mass and v is the velocity.
  • Potential Energy (PE): Stored energy that has the potential to be converted into other forms of energy. There are different types of potential energy:
    • Gravitational Potential Energy (GPE): Energy due to an object's position relative to a gravitational field. GPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above a reference point.
    • Elastic Potential Energy (EPE): Energy stored in a deformable object, such as a spring, when it is stretched or compressed. EPE = 1/2 kx², where k is the spring constant and x is the displacement from the equilibrium position.
• Work-Energy Theorem: This theorem states that the net work done on an object is equal to the change in its kinetic energy. W_net = ΔKE = KE_final - KE_initial.
• Conservation of Energy: In a closed system, the total energy remains constant; it can be transformed from one form to another, but it cannot be created or destroyed.
• Power (P): The rate at which work is done or energy is transferred. P = W/t = ΔE/t, where t is the time. Power is measured in Watts (W).

Constructing Work and Energy Diagrams

A work and energy diagram is a visual representation of the energy transformations occurring in a system. It typically includes:

1. System Definition: Clearly define the system you are analyzing. This is crucial as it determines which forces are internal and external to the system.
2. Initial State: Identify the forms of energy present in the system at the beginning of the process. This could include kinetic energy, gravitational potential energy, elastic potential energy, or internal energy (thermal energy).
3. Energy Transfers (Work): Identify any external forces doing work on the system. This could be positive work (energy added to the system) or negative work (energy removed from the system). Represent these as arrows entering or leaving the system boundary.
4. Final State: Identify the forms of energy present in the system at the end of the process.
5. Dissipated Energy (Thermal Energy): Identify any energy that is converted into thermal energy due to friction or other dissipative forces. This is often represented as an arrow leaving the system, indicating energy loss.
6. Diagram Representation: The diagram usually consists of a circle or box representing the system, with arrows indicating energy flow. The arrows should be labeled with the type of energy and the amount (if known).

Example Problems and Solutions

Let's illustrate the use of work and energy diagrams with several example problems:

Problem 1: A Block Sliding Down an Inclined Plane

A 2 kg block slides down a frictionless inclined plane that is 3 meters high. What is the speed of the block at the bottom of the incline?

Solution:

1. System Definition: The system is the block and the Earth.
2. Initial State: At the top of the incline, the block has gravitational potential energy (GPE = mgh). Since it starts from rest, its kinetic energy is zero.
3. Energy Transfers (Work): Since the inclined plane is frictionless, there is no work done by friction.
4. Final State: At the bottom of the incline, the block has kinetic energy (KE = 1/2 mv²). Its gravitational potential energy is zero (assuming the bottom of the incline is our reference point for h=0).
5. Dissipated Energy: No energy is dissipated due to friction.

Work and Energy Diagram:

• Initial State: GPE (mgh) --> System --> Final State: KE (1/2 mv²)

Energy Conservation Equation:

GPE_initial = KE_final

mgh = 1/2 mv²

(2 kg)(9.8 m/s²)(3 m) = 1/2 (2 kg) v²

v² = 58.8 m²/s²

v = 7.67 m/s

Answer: The speed of the block at the bottom of the incline is 7.67 m/s.

Problem 2: A Spring-Mass System

A 0.5 kg mass is attached to a spring with a spring constant of 50 N/m. The spring is compressed by 0.2 meters and then released. What is the maximum speed of the mass?

Solution:

1. System Definition: The system is the mass and the spring.
2. Initial State: When the spring is compressed, the system has elastic potential energy (EPE = 1/2 kx²). The mass is initially at rest, so its kinetic energy is zero.
3. Energy Transfers (Work): No external work is done on the system.
4. Final State: At the point of maximum speed, all the elastic potential energy has been converted into kinetic energy (KE = 1/2 mv²). The spring is at its equilibrium position (x=0), so its elastic potential energy is zero.
5. Dissipated Energy: We assume there is no friction.

Work and Energy Diagram:

• Initial State: EPE (1/2 kx²) --> System --> Final State: KE (1/2 mv²)

Energy Conservation Equation:

EPE_initial = KE_final

1/2 kx² = 1/2 mv²

1/2 (50 N/m)(0.2 m)² = 1/2 (0.5 kg) v²

v² = 4 m²/s²

v = 2 m/s

Answer: The maximum speed of the mass is 2 m/s.

Problem 3: A Block Pulled by a Force with Friction

A 5 kg block is pulled across a horizontal surface by a force of 20 N at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the block and the surface is 0.2. If the block starts from rest, what is its speed after it has been pulled 2 meters?

Solution:

1. System Definition: The system is the block.
2. Initial State: The block starts from rest, so its kinetic energy is zero.
3. Energy Transfers (Work): The applied force does positive work on the block. Friction does negative work on the block.
4. Final State: The block has kinetic energy (KE = 1/2 mv²).
5. Dissipated Energy: Work is dissipated by friction as thermal energy.

Calculating the Forces:

• Applied Force in the horizontal direction: F_x = F cos θ = 20 N * cos(30°) = 17.32 N
• Normal Force: The normal force (N) is the force exerted by the surface perpendicular to the block. It is equal to the weight of the block minus the vertical component of the applied force: N = mg - F_y = (5 kg)(9.8 m/s²) - 20 N * sin(30°) = 49 N - 10 N = 39 N
• Frictional Force: f_k = μ_k * N = 0.2 * 39 N = 7.8 N

Calculating the Work Done:

• Work done by the applied force: W_applied = F_x * d = (17.32 N)(2 m) = 34.64 J
• Work done by friction: W_friction = -f_k * d = -(7.8 N)(2 m) = -15.6 J

Work and Energy Diagram:

• Initial State: 0 J --> System --> Final State: KE (1/2 mv²)
• Energy Transfers: W_applied (34.64 J) enters the system.
• Dissipated Energy: |W_friction| (15.6 J) leaves the system as thermal energy.

Work-Energy Theorem:

W_net = ΔKE

W_applied + W_friction = KE_final - KE_initial

34.64 J - 15.6 J = 1/2 (5 kg) v² - 0

19.04 J = 2.5 kg * v²

v² = 7.616 m²/s²

v = 2.76 m/s

Answer: The speed of the block after being pulled 2 meters is 2.76 m/s.

Problem 4: A Roller Coaster

A roller coaster car with a mass of 500 kg starts from rest at the top of a 30-meter high hill. It then goes down the hill and up a second hill that is 20 meters high. Assuming no friction, what is the speed of the car at the top of the second hill?

Solution:

1. System Definition: The system is the roller coaster car and the Earth.
2. Initial State: At the top of the first hill, the car has gravitational potential energy (GPE = mgh). Since it starts from rest, its kinetic energy is zero.
3. Energy Transfers (Work): No external work is done on the system since we are assuming no friction.
4. Final State: At the top of the second hill, the car has both gravitational potential energy (GPE = mgh) and kinetic energy (KE = 1/2 mv²).
5. Dissipated Energy: No energy is dissipated due to the assumption of no friction.

Work and Energy Diagram:

• Initial State: GPE_initial (mgh_1) --> System --> Final State: KE (1/2 mv²) + GPE_final (mgh_2)

Energy Conservation Equation:

GPE_initial = KE_final + GPE_final

mgh_1 = 1/2 mv² + mgh_2

(500 kg)(9.8 m/s²)(30 m) = 1/2 (500 kg) v² + (500 kg)(9.8 m/s²)(20 m)

147000 J = 250 kg * v² + 98000 J

49000 J = 250 kg * v²

v² = 196 m²/s²

v = 14 m/s

Answer: The speed of the car at the top of the second hill is 14 m/s.

Problem 5: A Ballistic Pendulum

A bullet of mass 0.01 kg is fired horizontally into a stationary wooden block of mass 1 kg suspended by a light string. The bullet becomes embedded in the block, and the block swings upward, rising a vertical distance of 0.2 meters. What was the initial speed of the bullet?

Solution:

This problem involves two steps: an inelastic collision and a conservation of energy problem.

Step 1: Inelastic Collision (Bullet embeds in the block)

1. System Definition: The system is the bullet and the block.
2. Initial State: The bullet has kinetic energy (KE = 1/2 mv²) and the block is at rest.
3. Energy Transfers (Work): Momentum is conserved in the collision, but kinetic energy is not conserved because it's an inelastic collision (some energy is converted to thermal energy during impact).
4. Final State: The bullet and block move together with a common velocity (v_f) immediately after the collision.

Momentum Conservation:

m_bullet * v_bullet + m_block * v_block = (m_bullet + m_block) * v_f

(0.01 kg) * v_bullet + (1 kg) * 0 = (0.01 kg + 1 kg) * v_f

0.01 * v_bullet = 1.01 * v_f

v_bullet = 101 * v_f (Equation 1)

Step 2: Conservation of Energy (Block swings upward)

1. System Definition: The system is the block (with the bullet inside) and the Earth.
2. Initial State: Immediately after the collision, the block and bullet have kinetic energy (KE = 1/2 (m_bullet + m_block) v_f²). We take the initial height as zero, so GPE = 0.
3. Energy Transfers (Work): No external work is done.
4. Final State: At the highest point of the swing, the block and bullet momentarily stop, so KE = 0. They have gained gravitational potential energy (GPE = (m_bullet + m_block)gh).
5. Dissipated Energy: Assuming air resistance is negligible, no energy is dissipated.

Energy Conservation Equation:

KE_initial = GPE_final

1/2 (m_bullet + m_block) * v_f² = (m_bullet + m_block) * g * h

1/2 (1.01 kg) * v_f² = (1.01 kg) * (9.8 m/s²) * (0.2 m)

1/2 * v_f² = 1.96 m²/s²

v_f² = 3.92 m²/s²

v_f = 1.98 m/s

Now, substitute v_f back into Equation 1:

v_bullet = 101 * v_f = 101 * (1.98 m/s) = 199.98 m/s ≈ 200 m/s

Answer: The initial speed of the bullet was approximately 200 m/s.

Common Mistakes and How to Avoid Them

• Incorrect System Definition: Choosing the wrong system can lead to incorrect identification of external forces and energy transfers. Carefully consider what is included in your system and what is external.
• Ignoring Dissipative Forces: Friction and air resistance are often present in real-world scenarios and must be accounted for. If the problem states "assume no friction," then you can ignore them. Otherwise, calculate the work done by friction and include it in your energy balance.
• Mixing Up Work and Energy: Remember that work is a transfer of energy, while energy is a property of the system. Do not treat them as interchangeable.
• Incorrectly Applying the Work-Energy Theorem: Make sure you are using the net work done on the object in the Work-Energy Theorem. This includes the work done by all forces acting on the object.
• Forgetting Potential Energy: Potential energy is often overlooked. Always consider gravitational and elastic potential energy when analyzing a system.
• Unit Conversions: Ensure that all quantities are expressed in consistent units (e.g., meters, kilograms, seconds) before performing calculations.

Advanced Applications

Work and energy diagrams can be extended to analyze more complex systems, including:

• Rotational Motion: Incorporate rotational kinetic energy (KE = 1/2 Iω², where I is the moment of inertia and ω is the angular velocity) and rotational work.
• Thermodynamics: Analyze heat transfer and internal energy changes in thermodynamic systems.
• Fluid Mechanics: Apply energy conservation principles to analyze fluid flow.
• Electrical Circuits: Analyze energy dissipation in resistors and energy storage in capacitors and inductors.

Conclusion

Work and energy diagrams are indispensable tools for visualizing and solving physics problems involving energy transformations. By carefully defining the system, identifying energy inputs and outputs, and applying the principles of work and energy, you can effectively analyze a wide range of physical phenomena. Mastering these diagrams will significantly enhance your understanding of energy concepts and improve your problem-solving skills. Remember to practice regularly and pay attention to the common mistakes discussed to develop a solid foundation in work and energy principles.