In Specialist Mathematics, the study of mechanics extends beyond forces and acceleration to include the scalar quantities of work and energy. This framework often provides a more efficient method for solving problems involving speed and position without requiring the direct calculation of time.
Work (\(W\)) is defined as the product of the component of a force in the direction of displacement and the magnitude of that displacement. It is a scalar quantity measured in Joules (J), where \(1 \text{ J} = 1 \text{ N m}\).
If a constant force \(\mathbf{F}\) acts on a particle while it undergoes a displacement \(\mathbf{d}\), the work done is given by the dot product:
\$\(W = \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}||\mathbf{d}| \cos(\theta)\)\$
where \(\theta\) is the angle between the force vector and the displacement vector.
For a force \(F(x)\) acting in the direction of the \(x\)-axis, the work done in moving a particle from \(x = a\) to \(x = b\) is the integral of the force with respect to displacement:
\$\(W = \int_{a}^{b} F(x) \, dx\)\$
EXAM TIP: When calculating work done against gravity on an inclined plane, remember that only the component of the weight parallel to the plane (or the vertical displacement) matters. \(W = mgh\), where \(h\) is the vertical height gained.
Kinetic Energy (\(K\)) is the energy possessed by an object due to its motion.
For a particle of mass \(m\) moving with speed \(v\):
\$\(K = \frac{1}{2}mv^2\)\$
Since \(v^2 = \mathbf{v} \cdot \mathbf{v}\), kinetic energy is always a non-negative scalar.
The change in kinetic energy of a particle is equal to the net work done on it by all forces acting upon it:
\$\(W_{net} = \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)\$
Where:
* \(u\) is the initial speed.
* \(v\) is the final speed.
KEY TAKEAWAY: The Work-Energy Principle is a rearrangement of Newton’s Second Law (\(F=ma\)). It is particularly useful when you are given forces and distances but the acceleration is not constant or time is not required.
Potential Energy (\(U\)) is the stored energy of a system due to its configuration or position.
In a uniform gravitational field (near the surface of the Earth), the potential energy of a mass \(m\) at a height \(h\) relative to a chosen datum level (\(h=0\)) is:
\$\(U_g = mgh\)\$
* Work done by gravity: \(W_g = -\Delta U_g\)
* Work done against gravity: \(W_{ext} = \Delta U_g\)
For a spring or elastic string with stiffness \(k\) and extension/compression \(x\):
\$\(F = kx\)\$
The work done in stretching the spring from \(x_1\) to \(x_2\) (which is the energy stored) is:
\$\(U_s = \int_{x_1}^{x_2} kx \, dx = \left[ \frac{1}{2}kx^2 \right]_{x_1}^{x_2}\)\$
\$\(U_s = \frac{1}{2}kx^2 \text{ (relative to the natural length)}\)\$
COMMON MISTAKE: Students often forget that elastic potential energy \(U_s = \frac{1}{2}kx^2\) is always positive, whether the spring is compressed or extended. However, gravitational potential energy can be negative depending on where you set your zero-reference (datum) line.
In a conservative system (where only conservative forces like gravity or spring forces do work), the total mechanical energy remains constant.
If non-conservative forces (like friction or air resistance) are present, the mechanical energy is not conserved. The work done by friction (\(W_f\)) results in a loss of mechanical energy:
\$\(E_{initial} + W_{other} = E_{final}\)\$
Usually written as:
\$\((K_i + U_i) - \text{Work done against friction} = (K_f + U_f)\)\$
| Force Type | Conservative? | Examples |
|---|---|---|
| Gravity | Yes | Weight (\(mg\)) |
| Elastic | Yes | Spring force (\(kx\)) |
| Friction | No | Kinetic friction, Air resistance |
| Applied | No | Pushing or pulling a block |
VCAA FOCUS: Problems involving a particle sliding down a curved track or a pendulum often require the Conservation of Energy. Because the Normal force is always perpendicular to the direction of motion, it does zero work, allowing you to use \(mgh_1 = \frac{1}{2}mv^2 + mgh_2\).
Power (\(P\)) is the rate at which work is done or the rate at which energy is transferred. It is a scalar measured in Watts (W), where \(1 \text{ W} = 1 \text{ J/s}\).
For a constant force \(\mathbf{F}\) acting on an object moving with velocity \(\mathbf{v}\):
\$\(P = \mathbf{F} \cdot \mathbf{v}\)\$
If the force and velocity are in the same direction:
\$\(P = Fv\)\$
If a vehicle of mass \(m\) moves at a constant velocity \(v\), the driving force \(D\) must equal the resistive forces \(R\). The power required is:
\$\(P = Dv = Rv\)\$
REMEMBER: If a car is accelerating, the driving force \(D\) is greater than the resistance \(R\). Use \(D - R = ma\) to find the driving force first, then use \(P = Dv\) to find the power at that specific instant.
| Concept | Formula | Notes |
|---|---|---|
| Work (Constant Force) | \(W = Fd \cos \theta\) | Scalar (Joules) |
| Work (Variable Force) | \(W = \int F(x) dx\) | Area under \(F-x\) graph |
| Kinetic Energy | \(K = \frac{1}{2}mv^2\) | Always \(\ge 0\) |
| Gravitational PE | \(U_g = mgh\) | \(h\) is vertical height |
| Elastic PE | \(U_s = \frac{1}{2}kx^2\) | \(x\) is extension/compression |
| Power | \(P = Fv\) | Rate of work done (Watts) |
STUDY HINT: Always start energy problems by defining a “zero” height for potential energy. This simplifies the \(mgh\) terms. Usually, the lowest point in the motion is the best choice for \(h=0\).
