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A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.?
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A block whose mass is 2kg fastened to a spring whose spring constant i...
Given:
- Mass of the block (m) = 2 kg
- Spring constant (k) = 100 N/m
- Displacement from the mean position (x) = 0.05 m
To find:
- Kinetic energy of the block when it is 0.05m away from its mean position
Formula:
The potential energy stored in a spring is given by the formula:
Potential Energy (PE) = (1/2)kx^2
The total mechanical energy of the block-spring system is conserved and remains constant throughout the motion. Therefore, the initial potential energy will be converted into kinetic energy when the block is at a distance of 0.05m from its mean position.
Calculation:
1. Calculate the potential energy at the mean position (x = 0):
PE = (1/2)kx^2
= (1/2)(100 N/m)(0 m)^2
= 0 J
2. Calculate the potential energy at x = 0.1 m:
PE = (1/2)kx^2
= (1/2)(100 N/m)(0.1 m)^2
= 0.5 J
3. The total mechanical energy at x = 0.1 m is equal to the initial potential energy:
Total Mechanical Energy (E) = PE
= 0.5 J
4. The total mechanical energy is the sum of kinetic energy (KE) and potential energy (PE). At x = 0.05 m, the potential energy is zero, so the total mechanical energy is equal to the kinetic energy:
E = KE
0.5 J = KE
Therefore, the kinetic energy of the block when it is 0.05m away from its mean position is 0.5 J.
Summary:
The kinetic energy of the block when it is 0.05m away from its mean position is calculated to be 0.5 J. The total mechanical energy of the block-spring system is conserved, and the initial potential energy is converted into kinetic energy when the block is displaced from its mean position.
- Mass of the block (m) = 2 kg
- Spring constant (k) = 100 N/m
- Displacement from the mean position (x) = 0.05 m
To find:
- Kinetic energy of the block when it is 0.05m away from its mean position
Formula:
The potential energy stored in a spring is given by the formula:
Potential Energy (PE) = (1/2)kx^2
The total mechanical energy of the block-spring system is conserved and remains constant throughout the motion. Therefore, the initial potential energy will be converted into kinetic energy when the block is at a distance of 0.05m from its mean position.
Calculation:
1. Calculate the potential energy at the mean position (x = 0):
PE = (1/2)kx^2
= (1/2)(100 N/m)(0 m)^2
= 0 J
2. Calculate the potential energy at x = 0.1 m:
PE = (1/2)kx^2
= (1/2)(100 N/m)(0.1 m)^2
= 0.5 J
3. The total mechanical energy at x = 0.1 m is equal to the initial potential energy:
Total Mechanical Energy (E) = PE
= 0.5 J
4. The total mechanical energy is the sum of kinetic energy (KE) and potential energy (PE). At x = 0.05 m, the potential energy is zero, so the total mechanical energy is equal to the kinetic energy:
E = KE
0.5 J = KE
Therefore, the kinetic energy of the block when it is 0.05m away from its mean position is 0.5 J.
Summary:
The kinetic energy of the block when it is 0.05m away from its mean position is calculated to be 0.5 J. The total mechanical energy of the block-spring system is conserved, and the initial potential energy is converted into kinetic energy when the block is displaced from its mean position.
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A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.?
Question Description
A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.?.
A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.?.
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A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.?, a detailed solution for A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.? has been provided alongside types of A block whose mass is 2kg fastened to a spring whose spring constant is 100 N per meter. It is pulled to a distance of 0.1 metre from over a frictionless surface and is released at t=0. Calculate the kinetic energy of the block when it is 0.05m away from its mean position.? theory, EduRev gives you an
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