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#energy calculation

18 public questions tagged with this topic.

Torque on Magnetic Dipole and Potential Energy

The magnetic potential energy of a dipole with \( m = 0.9 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \(

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). U_m = -m B cosθ . Given: m = 0.9 A m² , B = 0.4 T , θ = 90° , cos 90° = 0 . Substitute: U_m = -0.9 × 0.4 × 0 = 0 J . Substituting values gives 0 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

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Torque on Magnetic Dipole and Potential Energy

The magnetic potential energy of a dipole with \( m = 0.6 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \(

**Torque on magnetic dipole** in uniform field B is τ = m × B, magnitude τ = m B sinθ, m moment (A·m²), B field (T), θ angle between m and B (degrees). Torque tends to align moment with field, zero at θ = 0°, maximum mB at 90°, direction given by right-hand rule. U_m = -m B cosθ . Given: m = 0.6 A m² , B = 0.3 T , θ = 60° , cos 60° = 0.5 . Substitute: U_m = -0.6 × 0.3 × 0.5 = -0.09 J . Substituting values gives -0.09 J, which matches expected magnitude for this magnetic

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

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Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 0.4 \, \text{kg}, k = 160 \, \text{N/m}, A = 7 \, \text{cm} \). What is the potential energy

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Potential energy: U = (1/2) k x² . k = 160 N/m, x = 0.035 m . U = 0.5 × 160 × (0.035)² = 0.5 × 160 × 0.001225 = 0.098 J .

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

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Energy in SHM - Kinetic, Potential and Total

A mass of \( 0.8 \, \text{kg} \) on a spring with \( k = 200 \, \text{N/m} \) has \( A = 10 \, \text{cm} \). What is the

**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 energy: E = (1/2) k A² = 0.5 × 200 × (0.1)² = 1 J . Potential energy: U = (1/2) k x² = 0.5 × 200 × (0.05)² = 0.25 J . Kinetic energy: K = E - U = 1 - 0.25 = 0.75 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and

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

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Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 1.2 \, \text{kg}, k = 480 \, \text{N/m}, A = 6 \, \text{cm} \). What is the potential energy

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Potential energy: U = (1/2) k x² . k = 480 N/m, x = 0.03 m . U = 0.5 × 480 × (0.03)² = 0.5 × 480 × 0.0009 = 0.216 J .

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

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Work Heat Distinction and Energy Transfer Modes

How much heat is required to vaporize 0.9 g of water at 100^circ C and 1 atm ? (Latent heat = 2256 J/g )

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). Δ Q = m L . m = 0.9 , L = 2256 . Δ Q = 0.9 × 2256 = 2030.4 J ≈ 2030 J . 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 > Work Heat Distinction and Energy Transfer Modes

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Heat Transfer Work Distinction and Internal Energy Change

How much heat is required to raise the temperature of 0.5 kg of carbon from 25^circ C to 55^circ C ? (Specific heat of c

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. Δ Q = m s Δ T . m = 0.5 , s = 600 , Δ T = 55 - 25 = 30 . Δ Q = 0.5 × 600 × 30 = 9000 J . Using first law ΔU = Q

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

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Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 790 J of heat and has 310 J of work done on it. What is the change in internal energy?

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. First Law: Δ Q = Δ U + Δ W . Δ Q = -790 (heat released), Δ W = -310 (work on system). -790 = Δ U - 310 ⇒ Δ U = -790 + 310 = -480 J . Using first law ΔU = Q - W, W = ∫ P

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

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First Law of Thermodynamics Applications

0.1 kg of a substance absorbs 1200 J of heat, increasing its temperature from 20°C to 50°C. What is its specific heat ca

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Specific heat: s = (Δ Q)/(m Δ T) . Given Δ Q = 1200 J , m = 0.1 kg , Δ T = 50 - 20 = 30 K . s = (1200)/(0.1 × 30) = 400 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV,

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

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Orbital Velocity and Time Period

How much energy is required to move a 300kg satellite from 7RE to 14RE from Earth’s center? (ME\=6×1024kg,RE\=6.4×106m,G

ΔE = −GMEm(1r2−1r1). r1 = 4.48×107m, r2 = 8.96×107m. ΔE = −6.67×10−11×6×1024×300(18.96×107−14.48×107). ΔE = −1.201×1017(−1.116×10−8)≈1.34×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.3 × 10⁹ J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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