Work and Energy 功和能

“It is important to realize that in physics today, we have no knowledge of what energy is.” Richard Feynman (1918 – 1988) Nobel prize winner

“It is important to realize that in physics today, we have no knowledge of what energy is.” Richard Feynman (1918 – 1988) Nobel prize winner Read: “Surely You’re Joking, Mr Feynman!”

We may not know what energy is, but we do know how to calculate it.

System and surroundings System Surroundings

The energy principle System Surroundings The change in energy of a system (E sys ) is equal to the work done by the surroundings (W surr ), and to other kinds of energy transfers between system and surroundings.

The total energy of the universe is conserved. System

Energy is transferred between system and surroundings. Never created, never destroyed. So for any choice of system:

Energy of a single particle Units of energy: the joule (J). 1 J = 1 kg (m/s) 2.

For a particle at rest, γ = 1. A particle not moving (v = 0) has energy proportional to its mass. Mass and energy are the same thing!

Mass turning into energy.

For a moving particle, γ > 1 E particle mc 2 K γmc 2 Total energy is the rest energy plus an extra amount, related to the particle’s motion.

For a moving particle, γ > 1 Rest energy E particle mc 2 K γmc 2 Kinetic energy 动能

Kinetic energy E particle mc 2 K γmc 2

Kinetic energy at low speeds When v << c, we can use a binomial expansion to simplify the expression for kinetic energy:

Kinetic energy at low speeds E particle mc 2 K γmc 2

Kinetic energy at low speeds E particle mc 2 K γmc 2

Work

Change in momentum is related to impulse: force multiplied by time.

What is related to force multiplied by distance?

Change in energy. This quantity is called “work” ( 功 ).

Impulse = Force x time Work = Force x distance …changes momentum …is a vector ( 矢量 ) …changes energy …is a scalar ( 标量 )

Work done by a constant force Units of work: the joule (J). Work is energy.

θ 1.0 m x y Example: How much work is done by the force F on the red block?

Only the component of the force in the direction of motion does work.

The vector dot product For any two vectors A and B: θ

Work done by a constant force where

Δx = +2 m F x = +3 N Is the work done on the block positive (+) or negative (-) ? The block is speeding up, so the work is positive.

Δx = +2 m F x = -3 N Is the work done on the block positive (+) or negative (-) ? The block is slowing down, so the work is negative.

Δx = -2 m F x = -3 N Is the work done on the block positive (+) or negative (-) ? The block is speeding up, so the work is positive.

Δx = -2 m F x = -3 N Is the work done on the block positive (+) or negative (-) ? The block is slowing down, so the work is negative.

Work done by a changing force