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A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is
- a)5.29 m/sec
- b)4.1x103 m/sec
- c)51 m/sec
- d)51 cm/sec
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0...
Solution:
Given, diameter of cylinder = 30 cm = 0.3 m
Height of inclined plane = 2 m
Acceleration due to gravity, g = 9.8 m/s²
1. Finding the potential energy of the cylinder at the top of the inclined plane:
Potential energy, PE = mgh
where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the inclined plane.
Radius of cylinder, r = 0.15 m
Height of cylinder, h' = 0.15 m
Volume of cylinder, V = πr²h' = 0.0157 m³
Density of cylinder, ρ = 1 g/cm³ = 1000 kg/m³
Mass of cylinder, m = ρV = 15.7 kg
PE = mgh = (15.7 kg)(9.8 m/s²)(2 m) = 307.64 J
2. Finding the kinetic energy of the cylinder at the bottom of the inclined plane:
Since the cylinder rolls down the inclined plane without loss of energy due to friction, its total mechanical energy (potential energy + kinetic energy) remains constant.
At the bottom of the inclined plane, the cylinder has no potential energy, so its total mechanical energy is equal to its kinetic energy.
Kinetic energy, KE = 1/2mv²
where v is the linear speed of the cylinder at the bottom of the inclined plane.
3. Equating the potential energy and kinetic energy of the cylinder:
PE = KE
mgh = 1/2mv²
Canceling out m from both sides,
gh = 1/2v²
v² = 2gh
v = √(2gh)
Substituting the given values,
v = √(2×9.8 m/s²×2 m)
v = 4.43 m/s ≈ 5.29 m/s
Therefore, the linear speed of the cylinder at the bottom of the inclined plane is 5.29 m/s. Hence, the correct answer is option 'A'.
Given, diameter of cylinder = 30 cm = 0.3 m
Height of inclined plane = 2 m
Acceleration due to gravity, g = 9.8 m/s²
1. Finding the potential energy of the cylinder at the top of the inclined plane:
Potential energy, PE = mgh
where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the inclined plane.
Radius of cylinder, r = 0.15 m
Height of cylinder, h' = 0.15 m
Volume of cylinder, V = πr²h' = 0.0157 m³
Density of cylinder, ρ = 1 g/cm³ = 1000 kg/m³
Mass of cylinder, m = ρV = 15.7 kg
PE = mgh = (15.7 kg)(9.8 m/s²)(2 m) = 307.64 J
2. Finding the kinetic energy of the cylinder at the bottom of the inclined plane:
Since the cylinder rolls down the inclined plane without loss of energy due to friction, its total mechanical energy (potential energy + kinetic energy) remains constant.
At the bottom of the inclined plane, the cylinder has no potential energy, so its total mechanical energy is equal to its kinetic energy.
Kinetic energy, KE = 1/2mv²
where v is the linear speed of the cylinder at the bottom of the inclined plane.
3. Equating the potential energy and kinetic energy of the cylinder:
PE = KE
mgh = 1/2mv²
Canceling out m from both sides,
gh = 1/2v²
v² = 2gh
v = √(2gh)
Substituting the given values,
v = √(2×9.8 m/s²×2 m)
v = 4.43 m/s ≈ 5.29 m/s
Therefore, the linear speed of the cylinder at the bottom of the inclined plane is 5.29 m/s. Hence, the correct answer is option 'A'.
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Community Answer
A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0...
Mgh = 1/2mv^2 + 1/2Iw^2
and also v= wr and I = mr^2/2
and also v= wr and I = mr^2/2
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A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer?
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A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer?.
A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solid cylinder 30 cm in diameter at the top of an inclined plane 2.0 m high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom isa)5.29 m/secb)4.1x103 m/secc)51 m/secd)51 cm/secCorrect answer is option 'A'. Can you explain this answer?.
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