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#total energy

9 public questions tagged with this topic.

Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.8 \, \text{kg}, k = 320 \, \text{N/m} \). If displaced by \( 5 \, \text{cm} \), what i

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Total energy: E = (1/2) k A² . A = 0.05 m, k = 320 N/m . E = 0.5 × 320 × (0.05)² = 0.5 × 320 × 0.0025 = 0.4 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.4 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.4 \, \text{kg}, k = 160 \, \text{N/m} \). If displaced by \( 6 \, \text{cm} \), what i

**Spring-mass system** has period T = 2π√(m/k), frequency f = (1/2π)√(k/m), ω = √(k/m), where k spring constant (N/m) and m mass (kg). For parallel combination, k_eff = k₁ + k₂, series gives 1/k_eff = 1/k₁ + 1/k₂, affecting ω = √(k_eff/m) and T = 2π√(m/k_eff). Total energy: E = (1/2) k A² . A = 0.06 m, k = 160 N/m . E = 0.5 × 160 × (0.06)² = 0.5 × 160 × 0.0036 = 0.288 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.288 J

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.9 \, \text{kg}, k = 360 \, \text{N/m} \). If displaced by \( 4 \, \text{cm} \), what i

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Total energy: E = (1/2) k A² . A = 0.04 m, k = 360 N/m . E = 0.5 × 360 × (0.04)² = 0.5 × 360 × 0.0016 = 0.288 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.288 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 0.5 \, \text{kg}, k = 200 \, \text{N/m} \). If the amplitude is \( 4 \, \text{cm} \), wh

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Total energy: E = (1/2) k A² . A = 0.04 m, k = 200 N/m . E = 0.5 × 200 × (0.04)² = 0.5 × 200 × 0.0016 = 0.16 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.16 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Energy in SHM - Kinetic, Potential and Total

A mass of \( 3 \, \text{kg} \) is attached to a spring with \( k = 300 \, \text{N/m} \). If displaced by \( 15 \, \text{

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Total energy: E = (1/2) k A² . A = 0.15 m, k = 300 N/m . E = (1/2) × 300 × (0.15)² = 0.5 × 300 × 0.0225 = 3.375 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A²,

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

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Escape Velocity and Bound Systems

A 700kg satellite orbits Earth at 10RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10−

E = −GMEm2r. r = 10RE = 6.4×107m. E = −6.67×10−11×6×1024×7002×6.4×107. E = −2.801×10171.28×108≈−2.19×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -2.2 × 10⁹ J. 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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Orbital Velocity and Time Period

A 1000kg satellite orbits Earth at 16RE from the center. What is its total energy? (ME\=6×1024kg,RE\=6.4×106m,G\=6.67×10

E = −GMEm2r. r = 16RE = 1.024×108m. E = −6.67×10−11×6×1024×10002×1.024×108. E = −4.002×10172.048×108≈−1.95×109J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.9 × 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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