Chapter 6: Work, Energy and Power
Work and energy are among the most fundamental concepts in physics. The work-energy theorem provides a powerful alternative to Newton's laws for solving many problems.
Work Done by a Force
Work (W) is done when a force causes displacement in the direction of the force.
W = F × d × cos(theta)
where theta is the angle between the force vector and the displacement vector.
Key formulas
Work is a scalar quantity. SI unit: joule (J) = N m = kg m2 s-2
1 joule = work done when 1 N force moves object 1 m in the direction of force.
Kinetic Energy
Kinetic energy (KE) is the energy of motion:
KE = (1/2) m v2
Work-Energy Theorem
"The net work done on an object equals the change in its kinetic energy."
Wnet = Delta(KE) = (1/2) m v2 - (1/2) m u2
This is extremely useful: instead of finding acceleration and using kinematics, we can find final speed directly from work done.
Potential Energy
Potential energy (PE) is stored energy due to position or configuration.
Key formulas
Conservation of Mechanical Energy
In a system with only conservative forces (gravity, spring): total mechanical energy is conserved.
KE + PE = constant
or: (1/2)mv2 + mgh = constant at all points
This means: energy lost as KE = energy gained as PE, and vice versa.
Power
Power (P) is the rate of doing work:
P = W/t = F × v
SI unit: watt (W) = J/s. Also: 1 horsepower = 746 W.
Elastic and Inelastic Collisions
- Elastic collision: Both momentum AND kinetic energy are conserved. (e.g., billiard balls)
- Inelastic collision: Momentum is conserved but KE is NOT conserved. (e.g., clay hitting clay)
- Perfectly inelastic: Objects stick together after collision.
For 1D elastic collision: v1' = (m1-m2)/(m1+m2) × u1 and v2' = 2m1/(m1+m2) × u1 (if m2 at rest)
---
A 10 kg box is pushed 5 m along a frictionless surface with 20 N force at 60° to horizontal. Find work done.
W = F d cos(theta) = 20 × 5 × cos(60°) = 100 × 0.5 = 50 J
Find the speed of a 5 kg object after Wnet = 100 J is done, starting from rest.
W = Delta(KE) → 100 = (1/2)(5)v2 → v2 = 40 → v = 6.32 m/s
A 2 kg ball is released from height 5 m (g = 10 m/s2). Find speed at the bottom.
Energy conservation: mgh = (1/2)mv2 → gh = v2/2 → v = √(2gh) = √(100) = 10 m/s
A spring with k = 200 N/m is compressed 0.1 m. Find the elastic PE.
PE = (1/2) k x2 = (1/2)(200)(0.01) = 1 J
A motor lifts 500 kg by 10 m in 20 s (g = 10 m/s2). Find power output.
Work = mgh = 500 × 10 × 10 = 50,000 J; P = W/t = 50,000/20 = 2500 W = 2.5 kW
A 1500 kg car moves at 30 m/s. Find its kinetic energy.
KE = (1/2)(1500)(900) = 675,000 J = 675 kJ
In a perfectly inelastic collision, 2 kg mass at 6 m/s hits stationary 4 kg mass. Find final velocity.
By momentum conservation: m1 u1 = (m1 + m2) v → 2 × 6 = 6v → v = 2 m/s
---
Common mistakes
- Thinking a person holding a heavy object is doing work — no displacement means no work done (in physics).
- Forgetting that only the component of force ALONG the displacement does work.
- Confusing power (rate of work) with energy (total work done).
- Applying conservation of mechanical energy when friction is present — friction is non-conservative and removes mechanical energy.
Summary
Work = F d cos(theta). The work-energy theorem links net work to change in KE. Potential energy is stored energy (gravitational: mgh; elastic: (1/2)kx2). Mechanical energy is conserved for conservative forces. Power is the rate of doing work. In collisions, momentum is always conserved; kinetic energy is conserved only in elastic collisions.