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A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)?
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Analysis:
To solve this problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the system (ring and block) is conserved when the string becomes vertical.
Initial State:
In the initial state, the block B is at rest and the ring is attached to it. The system has potential energy due to the height of the block above the ground and no kinetic energy.
Final State:
When the string becomes vertical, the block B has fallen a distance equal to the length of the string. At this point, the block B has reached its maximum velocity and the ring has zero velocity.
Conservation of Mechanical Energy:
Since there is no external force acting on the system (neglecting air resistance), the total mechanical energy of the system is conserved. This means that the initial mechanical energy is equal to the final mechanical energy.
Initial Mechanical Energy:
The initial mechanical energy of the system is given by the potential energy of the block B due to its height above the ground. The potential energy is given by the equation: PE = mgh, where m is the mass of the block B, g is the acceleration due to gravity, and h is the height of the block above the ground.
Final Mechanical Energy:
The final mechanical energy of the system is given by the kinetic energy of the block B. The kinetic energy is given by the equation: KE = (1/2)mv^2, where m is the mass of the block B and v is its velocity.
Equating Initial and Final Mechanical Energies:
Setting the initial mechanical energy equal to the final mechanical energy, we have: mgh = (1/2)mv^2.
Simplifying the Equation:
Since the mass of the block B cancels out, we are left with: gh = (1/2)v^2.
Solving for Velocity:
To find the velocity v, we can rearrange the equation as follows: v^2 = 2gh.
Taking the square root of both sides: v = √(2gh).
Substituting Values:
Substituting the given values into the equation, we have: v = √(2 * 10 * 0.6) = √(12) = 3.46 m/s.
Final Answer:
The velocity of the ring when the string becomes vertical is 3.46 m/s.
To solve this problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the system (ring and block) is conserved when the string becomes vertical.
Initial State:
In the initial state, the block B is at rest and the ring is attached to it. The system has potential energy due to the height of the block above the ground and no kinetic energy.
Final State:
When the string becomes vertical, the block B has fallen a distance equal to the length of the string. At this point, the block B has reached its maximum velocity and the ring has zero velocity.
Conservation of Mechanical Energy:
Since there is no external force acting on the system (neglecting air resistance), the total mechanical energy of the system is conserved. This means that the initial mechanical energy is equal to the final mechanical energy.
Initial Mechanical Energy:
The initial mechanical energy of the system is given by the potential energy of the block B due to its height above the ground. The potential energy is given by the equation: PE = mgh, where m is the mass of the block B, g is the acceleration due to gravity, and h is the height of the block above the ground.
Final Mechanical Energy:
The final mechanical energy of the system is given by the kinetic energy of the block B. The kinetic energy is given by the equation: KE = (1/2)mv^2, where m is the mass of the block B and v is its velocity.
Equating Initial and Final Mechanical Energies:
Setting the initial mechanical energy equal to the final mechanical energy, we have: mgh = (1/2)mv^2.
Simplifying the Equation:
Since the mass of the block B cancels out, we are left with: gh = (1/2)v^2.
Solving for Velocity:
To find the velocity v, we can rearrange the equation as follows: v^2 = 2gh.
Taking the square root of both sides: v = √(2gh).
Substituting Values:
Substituting the given values into the equation, we have: v = √(2 * 10 * 0.6) = √(12) = 3.46 m/s.
Final Answer:
The velocity of the ring when the string becomes vertical is 3.46 m/s.
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A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)?
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A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)?.
A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small ring of mass 'm'is attached on one end of a light string of length l=0.6m. The other end of the string is tied to a small blockB of mass '2m' .The ring is free to move on smooth horizontal rod.The block B is released from rest from the position shown.The velocity of the ring when the string become vertical is(g=10m/s^2)?.
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