Work, Energy, and Power for the ESAT

Work Done and Force

When an external force acts on an object and causes it to move through a distance, energy is transferred. This process of energy transfer is defined as work. It is important to note that work is only done if there is a component of the distance moved along the same line as the force. If the displacement is in the same direction as the force, energy is transferred to the object. If the displacement is in the opposite direction, energy is removed from the object.

The mathematical definition of work is given by the product of the force and the distance moved in the direction of that force:

$W = Fs$

Where $W$ is work, $F$ is force, and $s$ is the distance component along the line of action of the force. While force and displacement are vectors, work is a scalar quantity. It can be positive or negative but has no direction. The SI unit for work is the joule ($J$). One joule is defined as the work done when a force of $1\text{ N}$ moves an object through a distance of $1\text{ m}$. In calculations, ensure units are consistent: energy in joules, force in newtons, and distance in metres.

When the force and the distance moved are not in the same direction, we use trigonometry to find the parallel component. If an object moves a distance $d$ at an angle $\theta$ to the force, the work done is:

$W = Fd \cos\theta$

Work as a Transfer of Energy

Work is essentially a mechanism for transferring energy. When work is done on an object, its energy increases. Conversely, when an object does work on its surroundings, its energy decreases. This is deeply linked to Newton's third law: if body A exerts a force on body B, body B exerts an equal and opposite force on body A. Since both move the same distance during contact, the energy lost by one is exactly equal to the energy gained by the other, provided no other forces are involved.

Consider two balls, A and B, colliding. If they exert a $60\text{ N}$ force on each other and move $5.0\text{ m}$ in the direction of A's motion, ball A does $300\text{ J}$ of work on ball B ($60 \times 5$). Ball B exerts a force in the opposite direction, doing $-300\text{ J}$ of work on A. Thus, $300\text{ J}$ is transferred from A to B with no net loss to the system.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. When an object of mass $m$ is raised through a vertical height $h$, the work done against gravity is converted into GPE:

$E_p = mgh$

To prove this, consider raising a mass at a constant speed. To prevent acceleration, the upward force $F$ must equal the weight $mg$. The work done over height $h$ is $W = F \times h = mgh$.

In physics problems, the 'zero point' for potential energy is arbitrary. You can define any height as $0\text{ J}$ (such as the floor or a table). What matters is the change in height, as the difference in energy between two points remains constant regardless of the reference level.

Worked Example: Loading a Trunk A trunk of mass $500\text{ kg}$ is pulled $4.0\text{ m}$ up a ramp to a height of $1.2\text{ m}$ by a $2000\text{ N}$ force.

1. Work done by the pulling force: $W = Fs = 2000 \times 4.0 = 8000\text{ J}$.
2. Increase in GPE: $mgh = 500 \times 10 \times 1.2 = 6000\text{ J}$.
3. Work against friction: $8000 - 6000 = 2000\text{ J}$.
4. Average friction force: $F = W/d = 2000 / 4.0 = 500\text{ N}$.

Kinetic Energy

Kinetic energy is the energy of an object in motion. For an object of mass $m$ moving at speed $v$, the kinetic energy is:

$E_k = \frac{1}{2}mv^2$

This formula can be derived by considering an object accelerated from rest by a force $F$ over distance $s$. Since $F = ma$ and $v^2 = u^2 + 2as$ (with $u=0$), we find $s = v^2 / 2a$. Substituting $a = F/m$ gives $s = mv^2 / 2F$. The work done is $W = Fs = F(mv^2 / 2F) = \frac{1}{2}mv^2$. This work is stored as the kinetic energy of the object.

Power

Power is defined as the rate at which energy is transferred or the rate at which work is done. It is calculated as:

$P = \frac{W}{t} = \frac{E}{t}$

The SI unit for power is the watt ($W$), where $1\text{ W} = 1\text{ J s}^{-1}$. For larger values, we use kilowatts ($kW$).

Note on the Kilowatt-hour (kWh): The kWh is a unit of energy, not power. It is the energy transferred by a $1\text{ kW}$ device in one hour. Since $1\text{ kW} = 1000\text{ J/s}$ and $1\text{ hour} = 3600\text{ s}$, then $1\text{ kWh} = 3,600,000\text{ J}$.

Worked Example: Car Power A car of mass $1200\text{ kg}$ accelerates from rest to $20\text{ ms}^{-1}$ in $8.0\text{ s}$.

1. Change in $E_k = \frac{1}{2} \times 1200 \times 20^2 = 240,000\text{ J}$.
2. Useful output power: $P = E/t = 240,000 / 8.0 = 30,000\text{ W}$ (or $30\text{ kW}$).

Conservation of Energy

The law of conservation of energy states that energy cannot be created or destroyed, only transferred between forms. In mechanical problems, we often track energy moving between 'active' and 'stored' forms.

• Active forms: Electrical, heat (thermal), light, kinetic, and sound.
• Stored forms: Chemical, gravitational potential, and strain (elastic) potential energy.

Worked Example: Oscillating Spring A $1.2\text{ kg}$ mass is attached to an unstretched spring and released. It falls $0.20\text{ m}$ to its lowest point.

1. Loss of GPE: $mgh = 1.2 \times 10 \times 0.20 = 2.4\text{ J}$.
2. Gain in strain energy: $\frac{1}{2}kx^2 = \frac{1}{2}k(0.20)^2 = 0.020k$.
3. By conservation, $0.020k = 2.4$, so $k = 120\text{ Nm}^{-1}$.

Useful Energy and Efficiency

Every energy transfer device has a intended purpose. Useful energy is the energy converted into the desired form, while wasted energy is converted into other forms (usually heat due to friction). For instance, in a lightbulb, light is useful and heat is wasted; in a heater, heat is useful and light is wasted.

The efficiency of a device is the ratio of useful output to total input:

$\text{percentage efficiency} = \frac{\text{useful output}}{\text{total input}} \times 100$

Efficiency can be calculated using either energy or power values, provided the units are consistent. Total output always equals the sum of useful and wasted output.

Worked Example: Heater Efficiency A heater draws $5.0\text{ A}$ from a $240\text{ V}$ supply and produces $960\text{ W}$ of thermal energy.

1. Total input power: $P = VI = 240 \times 5.0 = 1200\text{ W}$.
2. Efficiency: $(960 / 1200) \times 100 = 80\%$.