Kinetics of Particles: Work and Energy

Why it matters

Understanding the work-energy principle in particle kinetics is crucial for solving real-world engineering problems, such as calculating the energy required to move vehicles or machinery. It helps engineers design systems that are energy-efficient and safe.

Key ideas

• Work-Energy Principle: The work done by all forces acting on a particle equals the change in its kinetic energy.
• Kinetic Energy (KE): The energy a particle possesses due to its motion, given by KE = 1/2·m·v², where m is mass and v is speed.
• Work Done (W): For a constant force vector, its dot product with the displacement, W = F·d·cosθ, where F is force, d is displacement, and θ is the angle between force and displacement.
• Potential Energy (PE): The energy stored in a particle due to its position or configuration, such as gravitational potential energy PE = m·g·h, where g is acceleration due to gravity and h is height.
• Conservation of Energy: Total energy of an isolated system is conserved, including internal and thermal energy. Mechanical energy T + V is conserved only when the net work of nonconservative forces is zero. Friction can convert mechanical energy into thermal energy.

General work-energy relation

For a constant-mass particle in an inertial frame, W_net = ∫ ΣF · dr = T₂ − T₁. For conservative forces, W_c = −ΔV, giving T₁ + V₁ + W_nc = T₂ + V₂. Do not count gravity both as work and as a potential-energy change in the same balance. For a linear spring, V_s = kx²/2.

The expression Fd cos θ below assumes a constant force vector. With varying force or direction, use the path integral. The expression mgh assumes approximately uniform gravitational acceleration.

Formulas

• KE = 1/2·m·v²
  • KE: Kinetic Energy (Joules)
  • m: Mass (kg)
  • v: Speed (m/s)
• W = F·d·cosθ
  • W: Work Done (Joules)
  • F: Force (Newtons)
  • d: Displacement (meters)
  • θ: Angle between force and displacement (degrees)
• PE = m·g·h
  • PE: Potential Energy (Joules)
  • m: Mass (kg)
  • g: Acceleration due to gravity (9.81 m/s²)
  • h: Height (meters)

Worked example

A 5 kg block travels 10 m up a straight 30° incline. A constant 50 N force acts parallel to the incline, and the block's speed increases from 2 m/s to 4 m/s. Find the increase in gravitational potential energy, the change in kinetic energy, and the friction required for these data to be consistent. Assume a stationary incline, constant sliding friction, and no other work-producing forces. Take g = 9.81 m/s².

The vertical rise is h = 10 sin 30° = 5 m. Therefore ΔV_g = mgh = 245.25 J. Gravity does −245.25 J of work; “work against gravity” is the positive potential-energy increase.

The applied force does W_P = 50 × 10 = 500 J because its direction is parallel to displacement. The incline angle is not the angle between this force and displacement.

The kinetic-energy change is ΔT = (1/2) × 5 × (4² − 2²) = 30 J.

The normal reaction does zero work. From W_P + W_g + W_f = ΔT:

W_f = 30 − 500 + 245.25 = −224.75 J.

Hence the constant opposing friction force is 224.75/10 = 22.475 N. Without friction or another energy-loss mechanism, the stated final speed would not be consistent with the applied force.

Common mistakes

• Confusing the angle in the work formula; ensure it's between the force and displacement.
• Forgetting to convert angles from degrees to radians when necessary.
• Ignoring the direction of forces when calculating work done.

For GATE ME

Questions often involve calculating work done, kinetic energy changes, and applying conservation of energy principles. Practice problems with varying forces, inclines, and initial conditions to master these concepts.

Quick check

1. What is the formula for kinetic energy?
2. How is work done calculated when force and displacement are perpendicular?
3. What remains constant in an isolated system according to the conservation of energy?

Answers: 1. KE = 1/2·m·v²; 2. Zero; 3. Total energy.