MechanicsHigh Yield (2 Qs in JEE)

Work, Power & Energy Formulas

Work-energy theorem, conservative forces, potential energy, elastic collisions in 1D/2D, and power.

7 Formulas•343+ Questions in Bank

Core Formulas & Mathematical Identities

W_{\text{net}} = W_{\text{conservative}} + W_{\text{non-conservative}} + W_{\text{external}} = \Delta K = K_f - K_i

Net work done by all forces on a particle equals the change in its kinetic energy.

Examiner Tip: Most reliable problem-solving tool in mechanics.

\vec{F} = -\vec{\nabla} U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right), \quad \Delta U = -W_{\text{cons}}

Force derived from scalar potential energy gradient.

Examiner Tip: Stable equilibrium occurs where dU/dx = 0 and d²U/dx² > 0 (minimum potential energy).

v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2, \quad e = \frac{v_2 - v_1}{u_1 - u_2} = 1

Final velocities after head-on perfectly elastic collision.

Examiner Tip: If m₁ = m₂ in elastic collision, velocities are completely exchanged: v₁ = u₂ and v₂ = u₁.

P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}, \quad P_{\text{avg}} = \frac{W_{\text{total}}}{\Delta t}

Rate of doing work (1 Watt = 1 Joule/second = 1 kg m² s⁻³).

Examiner Tip: 1 horsepower (hp) ≈ 746 Watts.

v_{\text{bottom}} \ge \sqrt{5gR}, \quad v_{\text{top}} \ge \sqrt{gR}, \quad T_{\text{bottom}} - T_{\text{top}} = 6mg, \quad T(\theta) = \frac{mv^2}{R} + mg\cos\theta

Critical thresholds for looping the vertical circle without slackening of string.

Examiner Tip: For a massless rigid rod, speed at top can be 0, so critical bottom speed is only √(4gR) = 2√(gR).

\Delta K = \frac{1}{2} \left(\frac{m_1 m_2}{m_1 + m_2}\right) (1 - e^2) (u_1 - u_2)^2, \quad v_1 = \frac{m_1 u_1 + m_2 u_2 - e m_2(u_1 - u_2)}{m_1 + m_2}

Loss of mechanical kinetic energy dissipated as heat and sound during collision with coefficient of restitution e.

Examiner Tip: For perfectly inelastic collision (e = 0), both bodies stick together with maximum possible KE loss.

U_s = \frac{1}{2} k x^2, \quad W_{\text{spring}} = -\Delta U_s = \frac{1}{2} k (x_i^2 - x_f^2)

Elastic potential energy stored in an ideal spring deformed by displacement x.

Examiner Tip: Effective spring constant for parallel springs: k_p = k₁ + k₂; for series springs: 1/k_s = 1/k₁ + 1/k₂.

Frequently tested misconceptions in Work, Power & Energy

•Work done by centripetal force is ALWAYS ZERO because force is perpendicular to instantaneous velocity (F · v = 0).
•In inelastic collisions, kinetic energy is NOT conserved (momentum is conserved).