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A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec?
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A body of mass 2kg slides down a curved track which is quadrant of a c...
By law of conservation of energy.mgh=1/2mv^2.v=√2gh.v=√2×9.8×1.v=√19.6.v=4.43m/s.
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A body of mass 2kg slides down a curved track which is quadrant of a c...
Solution:
Given:
- Mass of the body (m) = 2 kg
- Radius of the curved track (r) = 1 meter
To find:
- The speed of the body at the bottom of the track
Acceleration due to gravity:
The force acting on the body as it slides down the track is the gravitational force. The acceleration due to gravity (g) is given by:
- g = 9.8 m/s^2 (approximately)
Finding the height of the track:
The track is a quadrant of a circle, and the radius is given as 1 meter. The height of the track can be found using the formula for the length of an arc:
- L = rθ
Since the track is a quadrant (θ = π/2), the length of the track is:
- L = (1 meter) × (π/2) = π/2 meters
The potential energy at the start:
The gravitational potential energy (PE) of the body at the top of the track is given by:
- PE = mgh
Since the body starts from rest, its initial velocity (v) is zero. Therefore, the potential energy at the start is equal to the kinetic energy at the bottom of the track.
The potential energy at the bottom:
At the bottom of the track, the body has reached its maximum speed. The potential energy is converted into kinetic energy. The kinetic energy (KE) of the body at the bottom of the track is given by:
- KE = 1/2 mv^2
Using the principle of conservation of energy, we can equate the potential energy at the start to the kinetic energy at the bottom:
- PE = KE
- mgh = 1/2 mv^2
Simplifying the equation:
- gh = 1/2 v^2
- v^2 = 2gh
- v = √(2gh)
Substituting the known values:
- v = √(2 × 9.8 × π/2)
- v = √(9.8π)
- v ≈ 4.43 m/s
Therefore, the speed of the body at the bottom of the track is approximately 4.43 m/s.
Answer: a) 4.43 m/s
Given:
- Mass of the body (m) = 2 kg
- Radius of the curved track (r) = 1 meter
To find:
- The speed of the body at the bottom of the track
Acceleration due to gravity:
The force acting on the body as it slides down the track is the gravitational force. The acceleration due to gravity (g) is given by:
- g = 9.8 m/s^2 (approximately)
Finding the height of the track:
The track is a quadrant of a circle, and the radius is given as 1 meter. The height of the track can be found using the formula for the length of an arc:
- L = rθ
Since the track is a quadrant (θ = π/2), the length of the track is:
- L = (1 meter) × (π/2) = π/2 meters
The potential energy at the start:
The gravitational potential energy (PE) of the body at the top of the track is given by:
- PE = mgh
Since the body starts from rest, its initial velocity (v) is zero. Therefore, the potential energy at the start is equal to the kinetic energy at the bottom of the track.
The potential energy at the bottom:
At the bottom of the track, the body has reached its maximum speed. The potential energy is converted into kinetic energy. The kinetic energy (KE) of the body at the bottom of the track is given by:
- KE = 1/2 mv^2
Using the principle of conservation of energy, we can equate the potential energy at the start to the kinetic energy at the bottom:
- PE = KE
- mgh = 1/2 mv^2
Simplifying the equation:
- gh = 1/2 v^2
- v^2 = 2gh
- v = √(2gh)
Substituting the known values:
- v = √(2 × 9.8 × π/2)
- v = √(9.8π)
- v ≈ 4.43 m/s
Therefore, the speed of the body at the bottom of the track is approximately 4.43 m/s.
Answer: a) 4.43 m/s
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A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec?.
A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1meter. all the surfaces are frictionless. if the body starts from rest, its speed at the bottom of the track is: a) 4.43m/sec b)2m/sec c)0.5m/sec d)19.6m/sec?.
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