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#energy conservation

30 public questions tagged with this topic.

Forced Oscillations and Resonance

Which condition ensures that the total mechanical energy in an SHM system remains conserved during the motion?

**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. Total mechanical energy (kinetic + potential) is conserved in SHM when no external dissipative forces (e.g., friction) act, allowing energy to transform without loss. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Absence of dissipative forces follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Energy in SHM - Kinetic, Potential and Total

In an ideal SHM system, what occurs to the total mechanical energy as the particle moves from the mean position to an ex

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. Total mechanical energy in ideal SHM (no friction) is conserved, remaining constant as kinetic energy converts to potential energy during the motion. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It remains constant follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Heat Transfer Work Distinction and Internal Energy Change

A system in a cyclic process performs 300 J of work and rejects 200 J of heat. What is the heat absorbed?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For cyclic: Δ U = 0 , Q_net = W . Q_absorb - Q_reject = W . Q_absorb - 200 = 300 ⇒ Q_absorb = 500 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Specific Heat Capacity and Latent Heat

Which of the following correctly describes the First Law of Thermodynamics?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. The First Law ( Δ Q = Δ U + Δ W ) states that heat added equals the increase in internal energy plus work done by the system, a form of energy conservation. Option B is correct. Using first law ΔU = Q - W, W = ∫

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the primary implication of the First Law of Thermodynamics?

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. The First Law of Thermodynamics is a statement of energy conservation: Δ Q = Δ U + Δ W . It implies that the total energy supplied to a system (as heat) equals the increase in internal energy plus the work done by the system. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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First Law of Thermodynamics Applications

Why does the First Law of Thermodynamics allow some processes that the Second Law prohibits?

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. The First Law ensures energy conservation ( Δ Q = Δ U + Δ W ), permitting any energy-balanced process. The Second Law introduces directionality and efficiency limits (e.g., no 100% heat-to-work conversion), restricting feasible processes. Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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Adiabatic Processes and Gamma Determination

Which of the following statements is incorrect regarding the First Law of Thermodynamics?

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. The First Law ( Δ Q = Δ U + Δ W ) is a conservation of energy principle, not requiring equilibrium or constant temperature. Option C is incorrect as it imposes an unnecessary condition. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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Isobaric and Isothermal Processes Work Calculation

Which of the following statements correctly describes the First Law of Thermodynamics?

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. The First Law ( Δ Q = Δ U + Δ W ) is a conservation principle, stating heat added equals internal energy increase plus work done. Option D is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 -

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Isobaric and Isothermal Processes Work Calculation

Which of the following correctly characterizes an adiabatic process?

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. An adiabatic process involves no heat transfer ( Δ Q = 0 ), with changes in internal energy driven by work. Option D is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields No heat transfer,

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Work, Energy and Orbital Mechanics

A body is launched from Earth at 13.5km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (13.5)2−(11.2)2 = 182.25−125.44 = 56.81. vf = 56.81≈7.54km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.5 km/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Potential Energy of Mass Systems

A body is launched from Earth at 11.8km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (11.8)2−(11.2)2 = 139.24−125.44 = 13.8. vf = 13.8≈3.71km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.7 km/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Gravitational Potential and Field

Minimum speed to escape from 2RE from Earth’s center is? (g\=9.8m/s2,RE\=6.4×106m)

ve = 2gRE22RE = gRE. ve = 9.8×6.4×106 = 6.272×107. ve≈7.92×103m/s≈7.9km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.9 km/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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