The Work-Energy Theorem is one of the foundational cornerstones of classical mechanics in CBSE Class 11 Physics. While deriving the theorem for a constant force using kinematic equations (v² - u² = 2as) is straightforward, CBSE examiners routinely allocate 3 to 5 marks for the Calculus Derivation of the Work-Energy Theorem for a Variable Force. In this guide, we break down the mathematical proof, highlight the exact chain rule substitution examiners look for, and contrast conservative vs non-conservative forces.
Direct Answer Injection: What is the Work-Energy Theorem?
The Work-Energy Theorem states that the net work done by all external forces (conservative, non-conservative, and external) acting on a particle is equal to the change in its kinetic energy. Mathematically: W_net = ΔK = K_f - K_i = (1/2) * m * v_f² - (1/2) * m * v_i². For a 1D variable force, the work done is represented as the integral: W = ∫ F(x) dx from x_i to x_f.
1. The Complete Calculus Derivation (Step-by-Step)
Follow this exact chronological structure to secure full marks in school examinations and board evaluations:
Step 1: Expressing Instantaneous Kinetic Energy
Let a particle of mass m be moving along the x-axis under the influence of a position-dependent variable force F(x). Its instantaneous kinetic energy K is given by:
K = (1/2) * m * v^2Step 2: Differentiating Kinetic Energy with Respect to Time
Differentiating both sides with respect to time t, using the chain rule of differentiation:
dK/dt = (1/2) * m * d(v^2)/dt = (1/2) * m * (2 * v * dv/dt) = m * v * (dv/dt)Step 3: Applying Newton's Second Law
From Newton's Second Law of Motion, acceleration is a = dv/dt, and the net force is F = m * a = m * (dv/dt). Substituting F into the rate of change of kinetic energy:
dK/dt = F * vStep 4: Transforming Temporal Derivative to Spatial Derivative (The Chain Rule Step)
Since instantaneous velocity is defined as the rate of displacement v = dx/dt:
dK/dt = F * (dx/dt)
dK = F * dxExaminer Note: This infinitesimal relation dK = F dx demonstrates that the infinitesimal work done dW = F dx produces an identical infinitesimal change in kinetic energy dK.
Step 5: Integrating Between Initial and Final Boundary Limits
Integrate both sides as the body moves from initial position x_i (where velocity is v_i and kinetic energy is K_i) to final position x_f (where velocity is v_f and kinetic energy is K_f):
∫_{K_i}^{K_f} dK = ∫_{x_i}^{x_f} F(x) dx
[K]_{K_i}^{K_f} = ∫_{x_i}^{x_f} F(x) dx
K_f - K_i = W_netSubstituting K_i = (1/2) * m * v_i² and K_f = (1/2) * m * v_f²:
(1/2) * m * v_f^2 - (1/2) * m * v_i^2 = W_netThus, the net work done by the variable force equals the change in kinetic energy of the body. Q.E.D.
2. Alternate Method: Direct Spatial Integration of Force
Some CBSE textbooks derive the theorem by beginning directly with the definition of work done by a variable force:
W = ∫_{x_i}^{x_f} F dx- Substitute
F = m * a = m * (dv/dt):W = ∫_{x_i}^{x_f} m * (dv/dt) dx - By the chain rule,
(dv/dt) * dx = (dx/dt) * dv = v dv. - Change limits of integration from spatial bounds (x_i, x_f) to velocity bounds (v_i, v_f):
W = ∫_{v_i}^{v_f} m * v dv = m * [v² / 2]_{v_i}^{v_f} = (1/2) * m * v_f² - (1/2) * m * v_i² = ΔK.
3. Common Pitfalls & Examiner Marking Scheme Breakdown
| Derivation Milestone | Marks Allotted | Common Student Error (How Marks are Lost) |
|---|---|---|
| Statement of Theorem & K = ½mv² | 0.5 Mark | Omitting the initial verbal statement of the theorem |
| Chain Rule: dK/dt = mv(dv/dt) = F(dx/dt) | 1.0 Mark | Failing to explain how v = dx/dt converts the differential to dK = F dx |
| Integration with Exact Boundary Limits | 1.0 Mark | Performing indefinite integration without limits or + C constant |
| Final Conclusion (W = ΔK) | 0.5 Mark | Not defining K_f and K_i clearly in terms of velocities |
4. Frequently Asked Questions (FAQ)
Does the Work-Energy Theorem hold true in non-inertial frames of reference?
Yes, the Work-Energy Theorem is valid in non-inertial (accelerating) reference frames, provided the work done by pseudo forces (fictitious inertial forces, W_pseudo = -m * a_frame * d) is included in the net work summation.
What is the graphical interpretation of work done by a variable force?
On a Force vs Position (F-x) graph, the net work done by the force is represented by the total area under the F-x curve bounded between the initial and final position coordinates. Area above the position axis represents positive work, while area below represents negative work.
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