Work-Energy Principle

The work-energy principle states that the net work done on a particle equals the change in its kinetic energy. This provides a powerful alternative to using Newton's Second Law directly, especially when forces vary or when you need to find speeds without knowing the time.

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1 Work Done by a Constant Force

W = F d cos theta

where F is the magnitude of the force, d is the displacement, and theta is the angle between the force and the direction of motion.

• If theta = 0 (force in direction of motion): W = Fd.
• If theta = 90 degrees (force perpendicular to motion): W = 0.
• If theta = 180 degrees (force opposing motion): W = -Fd.

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2 The Work-Energy Theorem

Net work done = Change in kinetic energy

W_net = (1/2)mv^2 - (1/2)mu^2

If multiple forces act, the net work is the sum of the work done by each force.

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3 Worked Example 1

A 5 kg box is pushed 4 m along a smooth horizontal surface by a force of 20 N. It starts at rest. Find the final speed.

Solution:

W = Fd = 20 * 4 = 80 J

(1/2)(5)v^2 - 0 = 80

v^2 = 32, v = 4 sqrt(2) approximately 5.66 m s^(-1)

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4 Gravitational Potential Energy

The work done against gravity when raising a mass m through a height h is:

W = mgh

This equals the gain in gravitational potential energy (GPE).

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5 Conservation of Energy

For a system with only conservative forces (gravity, elastic):

KE + PE = constant

(1/2)mv^2 + mgh = constant

Worked Example 2

A ball of mass 0.5 kg is thrown upward at 10 m s^(-1). Find the maximum height (ignoring air resistance).

At the top, v = 0:

(1/2)(0.5)(10)^2 = (0.5)(9.8)(h)

25 = 4.9h, h = 5.10 m

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6 Power

Power is the rate of doing work:

P = dW/dt = Fv (when force is in the direction of motion)

The SI unit is the watt (W): 1 W = 1 J s^(-1).

Worked Example 3

A car of mass 1200 kg travels at a constant speed of 30 m s^(-1) on a level road against a resistance of 600 N. Find the power of the engine.