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A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a ...
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a ...
Given information:
- Mass of the block = 10 kg
- Velocity of the plank = 1 m/s
- Coefficient of friction between the plank and block = 0.4
Introduction:
In this scenario, a block is placed on a moving plank with an initial velocity of 1 m/s. The plank is on a frictionless horizontal ground surface. We need to determine the work done by the external force that maintains the constant speed of the plank.
Analysis:
1. When the block is placed on the plank, it experiences a frictional force due to the coefficient of friction between the plank and the block.
2. The frictional force opposes the relative motion between the block and the plank and acts in the opposite direction of the plank's velocity.
3. Since the plank is moving with a constant speed, the net force acting on the system (block + plank) is zero.
4. The frictional force can be calculated using the equation: frictional force = coefficient of friction × normal force.
5. The normal force acting on the block is equal to its weight, which can be calculated as: normal force = mass of block × acceleration due to gravity.
6. The acceleration due to gravity is approximately 9.8 m/s².
7. Substituting the given values, we can calculate the normal force and the frictional force.
Calculation:
- Mass of the block = 10 kg
- Acceleration due to gravity = 9.8 m/s²
Normal force = mass of block × acceleration due to gravity
Normal force = 10 kg × 9.8 m/s²
Normal force = 98 N
Frictional force = coefficient of friction × normal force
Frictional force = 0.4 × 98 N
Frictional force = 39.2 N
Work done:
The work done by the external force maintaining the constant speed of the plank is zero. This is because the frictional force does not cause any displacement in the direction of the force. The block and the plank move together as a system without any change in kinetic energy.
Conclusion:
The work done by the external force maintaining the constant speed of the plank is zero. This is because the frictional force between the plank and the block does not cause any displacement in the direction of the force. The block and the plank move together as a system without any change in kinetic energy.
- Mass of the block = 10 kg
- Velocity of the plank = 1 m/s
- Coefficient of friction between the plank and block = 0.4
Introduction:
In this scenario, a block is placed on a moving plank with an initial velocity of 1 m/s. The plank is on a frictionless horizontal ground surface. We need to determine the work done by the external force that maintains the constant speed of the plank.
Analysis:
1. When the block is placed on the plank, it experiences a frictional force due to the coefficient of friction between the plank and the block.
2. The frictional force opposes the relative motion between the block and the plank and acts in the opposite direction of the plank's velocity.
3. Since the plank is moving with a constant speed, the net force acting on the system (block + plank) is zero.
4. The frictional force can be calculated using the equation: frictional force = coefficient of friction × normal force.
5. The normal force acting on the block is equal to its weight, which can be calculated as: normal force = mass of block × acceleration due to gravity.
6. The acceleration due to gravity is approximately 9.8 m/s².
7. Substituting the given values, we can calculate the normal force and the frictional force.
Calculation:
- Mass of the block = 10 kg
- Acceleration due to gravity = 9.8 m/s²
Normal force = mass of block × acceleration due to gravity
Normal force = 10 kg × 9.8 m/s²
Normal force = 98 N
Frictional force = coefficient of friction × normal force
Frictional force = 0.4 × 98 N
Frictional force = 39.2 N
Work done:
The work done by the external force maintaining the constant speed of the plank is zero. This is because the frictional force does not cause any displacement in the direction of the force. The block and the plank move together as a system without any change in kinetic energy.
Conclusion:
The work done by the external force maintaining the constant speed of the plank is zero. This is because the frictional force between the plank and the block does not cause any displacement in the direction of the force. The block and the plank move together as a system without any change in kinetic energy.
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)??
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)??.
A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)??.
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