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#energy conservation

21 public questions tagged with this topic.

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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Gravitational Potential and Field

A projectile is launched at 3km/s from Earth’s surface. What is its maximum distance from the center? (Escape speed = 11

12vi2−ve22 = −ve22REr. 4.5−62.72 = −62.72REr. rRE = 62.7258.22≈1.077. r = 1.077×6.4×106≈6.89×106m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.9 × 10⁶ m. 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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Variation of g with Height and Depth

A body is launched from Earth at 14km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

vf2 = vi2−ve2. vf2 = (14)2−(11.2)2 = 196−125.44 = 70.56. vf = 70.56≈8.4km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.4 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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Energy Conservation and Friction

A 5kg block slides down a frictionless incline from 6m height. What is its speed at the bottom? (Take g\=10m/s2)

Potential energy mgh=5×10×6=300J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 11 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Energy Conservation and Friction

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Work, Energy and Laws of Motion - Mixed Practice

A 2.5kg mass falls from 8m onto a spring (k\=1000N/m). What is the maximum compression? (Take g\=10m/s2)

Potential energy mgh=2.5×10×8=200J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 200 J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Work, Energy and Laws of Motion - Mixed Practice

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Laws of Motion and Energy Conservation

A 8kg mass falls from 5m onto a spring (k\=2500N/m). What is the maximum compression? (Take g\=10m/s2)

Potential energy mgh=8×10×5=400J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 400 J. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Laws of Motion and Energy Conservation

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Two-Dimensional Motion and Vectors

In projectile motion, what is true about the total mechanical energy, assuming no air resistance?

Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields It remains conserved. Hence option C satisfies projectile formulas.

Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Two-Dimensional Motion and Vectors

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