Understanding the Work-Energy Theorem and Kinematic Equations

The question asks us to identify which standard equation relating initial speed ($\small u$), final speed ($\small v$), acceleration ($\small a$), and distance ($\small s$) is a special case of the work-energy theorem. The work-energy theorem is a fundamental principle in physics that connects work done on an object to its change in kinetic energy.

The Work-Energy Theorem

The work-energy theorem states that the net work done by all forces acting on an object is equal to the change in its kinetic energy. Mathematically, this is expressed as:

$\small W_{net} = \Delta KE$

Where:

• $\small W_{net}$ is the net work done on the object.
• $\small \Delta KE$ is the change in kinetic energy, which is the final kinetic energy minus the initial kinetic energy ($\small KE_{final} - KE_{initial}$).

The kinetic energy of an object with mass $\small m$ and speed $\small v$ is given by $\small KE = \frac{1}{2}mv^2$. So, the change in kinetic energy is:

$\small \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Here, $\small u$ is the initial speed and $\small v$ is the final speed.

Applying the Theorem to Constant Acceleration

Consider an object moving in one dimension under a constant net force $\small F$. If this force acts over a distance $\small s$ in the direction of motion, the work done by this force is:

$\small W = F \cdot s$

According to Newton's second law of motion, a constant net force $\small F$ acting on an object of constant mass $\small m$ produces a constant acceleration $\small a$, given by $\small F = ma$.

Substituting $\small F = ma$ into the work equation, we get the work done in terms of acceleration and distance:

$\small W = (ma)s$

Deriving the Kinematic Equation

Now, let's equate the work done to the change in kinetic energy according to the work-energy theorem:

$\small W_{net} = \Delta KE$

$\small (ma)s = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Assuming the mass $\small m$ is not zero, we can divide both sides of the equation by $\small m$:

$\small as = \frac{1}{2}v^2 - \frac{1}{2}u^2$

To eliminate the fraction, multiply both sides by 2:

$\small 2as = v^2 - u^2$

Rearranging the terms gives us a familiar kinematic equation:

$\small v^2 - u^2 = 2as$

This derivation shows that the equation $\small v^2 - u^2 = 2as$ is a direct consequence of the work-energy theorem when the net force (and thus acceleration) is constant and acts over a displacement in the same direction.

Comparing with the Options

We are given four options:

1. $\small v^2 - u^2 = 2as$
2. $\small v^2 \cdot u^2 = 2as$
3. $\small v^2 + u^2 = 2as$
4. $\small v^2 \cdot u = 2as$

Comparing the equation derived from the work-energy theorem ($\small v^2 - u^2 = 2as$) with the given options, we see that Option 1 matches exactly. The other options are not standard kinematic equations and cannot be derived from the work-energy theorem under the conditions of constant acceleration.

Therefore, the equation $\small v^2 - u^2 = 2as$ is a special case of the work-energy theorem.

Revision Table: Key Concepts

Let's quickly summarize the key concepts used:

• Work-Energy Theorem: Links work and energy change.
• Kinetic Energy: Energy an object has due to its motion.
• Work Done: Energy transferred by a force.
• Constant Acceleration: A condition where acceleration does not change over time.
• Kinematic Equation: A formula describing motion (position, velocity, acceleration, time) under constant acceleration.

Additional Information on Work-Energy Theorem and Kinematics

The equation $\small v^2 - u^2 = 2as$ is one of the key kinematic equations used for solving problems involving motion with constant acceleration. The full set of kinematic equations (for motion in one dimension with constant acceleration) includes:

• $\small v = u + at$ (relates final velocity, initial velocity, acceleration, and time)
• $\small s = ut + \frac{1}{2}at^2$ (relates displacement, initial velocity, acceleration, and time)
• $\small v^2 = u^2 + 2as$ or $\small v^2 - u^2 = 2as$ (relates final velocity, initial velocity, acceleration, and displacement)
• $\small s = \frac{(u+v)}{2}t$ (relates displacement, average velocity, and time)

While the work-energy theorem is a general principle, the derivation of $\small v^2 - u^2 = 2as$ from it requires the assumption of constant net force (leading to constant acceleration) and motion along a straight line where the force and displacement are parallel. This highlights how general principles like the work-energy theorem can simplify to specific formulas under certain conditions, making them applicable to particular types of problems in mechanics.