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A conveyor belt is moving at a constant speed of 2 metre per second a box is gently dropped on it the coefficient of frequency between them is to 0.5 the distance that the box will move relative to belt before coming to rest on it? ANSWER IS 0.4m. Any one can solve it?
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A conveyor belt is moving at a constant speed of 2 metre per second a ...
Calculating the Distance a Box will Move Relative to a Moving Conveyor Belt
Given:
- Speed of conveyor belt = 2 m/s
- Coefficient of friction between belt and box = 0.5
To Find:
- Distance that the box will move relative to belt before coming to rest on it
Formula:
- Friction force = Coefficient of friction x Normal force
- Normal force = Mass x Gravity
- Net force = Mass x Acceleration
- Net force = Applied force - Friction force
- Distance = (Initial velocity^2)/(2 x Acceleration)
Solution:
As the box is gently dropped on the conveyor belt, its initial velocity relative to the belt is zero. The box will come to rest when the friction force between the belt and the box is equal to the force of gravity acting on the box.
Let's assume that the mass of the box is 1 kg. The normal force acting on the box is:
- Normal force = Mass x Gravity
- Normal force = 1 kg x 9.81 m/s^2
- Normal force = 9.81 N
The friction force acting on the box is:
- Friction force = Coefficient of friction x Normal force
- Friction force = 0.5 x 9.81 N
- Friction force = 4.905 N
The net force acting on the box is:
- Net force = Applied force - Friction force
- Net force = Mass x Acceleration
- Acceleration = Net force / Mass
- Acceleration = (0 - 4.905 N) / 1 kg
- Acceleration = -4.905 m/s^2
The distance that the box will move relative to belt before coming to rest on it can be calculated using the formula:
- Distance = (Initial velocity^2)/(2 x Acceleration)
- Distance = (0^2)/(2 x -4.905 m/s^2)
- Distance = 0 m
So, the box will come to rest on the conveyor belt without moving relative to it. However, if the initial velocity of the box relative to the belt was not zero, then the distance that the box would move relative to belt before coming to rest on it can be calculated using the above formula.
For example, if the initial velocity of the box relative to the belt was 1 m/s, then the distance that the box would move relative to belt before coming to rest on it can be calculated as:
- Distance = (Initial velocity^2)/(2 x Acceleration)
- Distance = (1^2)/(2 x -4
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A conveyor belt is moving at a constant speed of 2 metre per second a ...
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A conveyor belt is moving at a constant speed of 2 metre per second a box is gently dropped on it the coefficient of frequency between them is to 0.5 the distance that the box will move relative to belt before coming to rest on it? ANSWER IS 0.4m. Any one can solve it?
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A conveyor belt is moving at a constant speed of 2 metre per second a box is gently dropped on it the coefficient of frequency between them is to 0.5 the distance that the box will move relative to belt before coming to rest on it? ANSWER IS 0.4m. Any one can solve it? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A conveyor belt is moving at a constant speed of 2 metre per second a box is gently dropped on it the coefficient of frequency between them is to 0.5 the distance that the box will move relative to belt before coming to rest on it? ANSWER IS 0.4m. Any one can solve it? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A conveyor belt is moving at a constant speed of 2 metre per second a box is gently dropped on it the coefficient of frequency between them is to 0.5 the distance that the box will move relative to belt before coming to rest on it? ANSWER IS 0.4m. Any one can solve it?.
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