Work–energy theorem Knowledge-Based Study

Study notes

1. The Fundamental Principle

   The work–energy theorem provides a direct relationship between the net work done on an object and the resulting change in its kinetic energy. In classical mechanics, whenever a net external force acts on an object over a displacement, it alters the object's speed.

   Mathematically, the theorem is stated as:

   $$W_{\text{net}} = \Delta K = K_f - K_i$$

   where:

   • $W_{\text{net}}$: Total (net) work performed by all forces acting on the object (measured in joules, J)
   • $K_i$: Initial kinetic energy, given by $K_i = \frac{1}{2}mv_i^2$
   • $K_f$: Final kinetic energy, given by $K_f = \frac{1}{2}mv_f^2$
   • $\Delta K$: Net change in kinetic energy ($K_f - K_i$)

   This theorem applies to both constant and variable forces, as well as one-dimensional and multi-dimensional motion.

   Connecting Newton's Second Law to Kinematics

   Consider an object of mass $m$ moving along a straight line under a constant net force $F_{\text{net}}$. According to Newton's second law:

   $$F_{\text{net}} = ma$$

   From the kinematic equations for constant acceleration, the relationship between initial velocity $v_i$, final velocity $v_f$, acceleration $a$, and displacement $d$ is:

   $$v_f^2 = v_i^2 + 2ad$$

   Solving for the product $ad$:

   $$ad = \frac{v_f^2 - v_i^2}{2}$$

   Now, compute the work done by the net force over displacement $d$:

   $$W_{\text{net}} = F_{\text{net}} d = (ma)d = m(ad)$$

   Substitute $ad$ into this expression:

   $$W_{\text{net}} = m \left( \frac{v_f^2 - v_i^2}{2} \right) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

   This directly establishes that net work equals the change in kinetic energy.

   Physical Interpretation of Work Signs

   Because kinetic energy depends on the square of speed, the sign of the net work dictates how the speed of the body changes:

   • Positive Net Work ($W_{\text{net}} > 0$): $K_f > K_i$. The net force has a component in the direction of motion, transferring energy into the object and causing it to speed up.
   • Negative Net Work ($W_{\text{net}} < 0$): $K_f < K_i$. The net force opposes the direction of motion, removing energy from the object and causing it to slow down.
   • Zero Net Work ($W_{\text{net}} = 0$): $K_f = K_i$. The object maintains a constant speed. This occurs when either the net force is zero or the net force acts perpendicular to the displacement at all times (such as in uniform circular motion).

   Two Equivalent Approaches

   When multiple forces act simultaneously on an object, $W_{\text{net}}$ can be determined in two distinct ways:

   1. Vector Sum Method: Find the net vector force first, $\vec{F}_{\text{net}} = \sum \vec{F}_k$, and then calculate work: $$W_{\text{net}} = \vec{F}_{\text{net}} \cdot \vec{d} = F_{\text{net}} d \cos\theta$$
   2. Scalar Sum Method: Calculate the work done by each individual force separately and sum them algebraically: $$W_{\text{net}} = W_1 + W_2 + W_3 + \dots + W_n$$

   Both methods yield the exact same scalar value for $W_{\text{net}}$. The scalar sum method is often preferable when dealing with combinations of conservative forces (like gravity) and non-conservative forces (like kinetic friction).

   Sample Problem: Braking Distance

   A car of mass $1200\text{ kg}$ travels at $20\text{ m/s}$. The driver brakes, and a constant friction force of $6000\text{ N}$ opposes the motion until the car stops.

   Solution:

   1. Identify kinetic energies: $$K_i = \frac{1}{2}(1200\text{ kg})(20\text{ m/s})^2 = 240\,000\text{ J}$$ $$K_f = 0\text{ J}$$
   2. Calculate $\Delta K$: $$\Delta K = 0 - 240\,000\text{ J} = -240\,000\text{ J}$$
   3. Express net work in terms of braking distance $d$: $$W_{\text{net}} = -f_k \cdot d = -(6000\text{ N}) \cdot d$$
   4. Apply the theorem: $$-6000 d = -240\,000 \implies d = 40\text{ m}$$

   Critical Conceptual Pitfall

   Remember that the theorem strictly equates $\Delta K$ to net work (the work of all forces combined), not the work done by an arbitrary single force. The work of a single force changes kinetic energy only if no other forces are doing work on the object.