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If a particle is moving once round a square c formed by the lines y = ± 1, x = ± 1 in the xy-plane with a force F = (x2 + xy + z)i + (x2 + y2 – z)j + xy k then the work done is ______
Correct answer is '0'. Can you explain this answer?
Most Upvoted Answer
If a particle is moving once round a square c formed by the lines y = ...
Analysis:
To find the work done by the force F on the particle as it moves once round the square C, we can use the definition of work done by a force: W = ∫F·dr, where F is the force and dr is the displacement vector along the path of the particle.
Path of the Particle:
The particle is moving once round a square formed by the lines y = 1 and x = 1. Since the lines y = 1 and x = 1 form the sides of the square, the path of the particle can be divided into four segments:
1. Segment 1: From (0,0,0) to (1,0,0)
2. Segment 2: From (1,0,0) to (1,1,0)
3. Segment 3: From (1,1,0) to (0,1,0)
4. Segment 4: From (0,1,0) to (0,0,0)
Calculating the Work Done:
We need to calculate the work done along each segment of the path and then sum them up to find the total work done.
1. Segment 1: Along this segment, only the x-component of the force F is non-zero, while the y and z components are zero. Therefore, the work done along this segment is zero.
2. Segment 2: Along this segment, both the x and y components of the force F are non-zero, while the z component is zero. Therefore, the work done along this segment is zero.
3. Segment 3: Along this segment, only the y-component of the force F is non-zero, while the x and z components are zero. Therefore, the work done along this segment is zero.
4. Segment 4: Along this segment, both the x and y components of the force F are zero, while the z component is non-zero. Therefore, the work done along this segment is zero.
Since the work done along each segment of the path is zero, the total work done is also zero.
Conclusion:
The correct answer is '0'. The work done by the force F on the particle as it moves once round the square C is zero. This is because the force components along each segment of the path are such that the work done along each segment is zero.
To find the work done by the force F on the particle as it moves once round the square C, we can use the definition of work done by a force: W = ∫F·dr, where F is the force and dr is the displacement vector along the path of the particle.
Path of the Particle:
The particle is moving once round a square formed by the lines y = 1 and x = 1. Since the lines y = 1 and x = 1 form the sides of the square, the path of the particle can be divided into four segments:
1. Segment 1: From (0,0,0) to (1,0,0)
2. Segment 2: From (1,0,0) to (1,1,0)
3. Segment 3: From (1,1,0) to (0,1,0)
4. Segment 4: From (0,1,0) to (0,0,0)
Calculating the Work Done:
We need to calculate the work done along each segment of the path and then sum them up to find the total work done.
1. Segment 1: Along this segment, only the x-component of the force F is non-zero, while the y and z components are zero. Therefore, the work done along this segment is zero.
2. Segment 2: Along this segment, both the x and y components of the force F are non-zero, while the z component is zero. Therefore, the work done along this segment is zero.
3. Segment 3: Along this segment, only the y-component of the force F is non-zero, while the x and z components are zero. Therefore, the work done along this segment is zero.
4. Segment 4: Along this segment, both the x and y components of the force F are zero, while the z component is non-zero. Therefore, the work done along this segment is zero.
Since the work done along each segment of the path is zero, the total work done is also zero.
Conclusion:
The correct answer is '0'. The work done by the force F on the particle as it moves once round the square C is zero. This is because the force components along each segment of the path are such that the work done along each segment is zero.
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Community Answer
If a particle is moving once round a square c formed by the lines y = ...
When the particle has moved once round any shape this means it has returned to the initial starting position. and since net displacement is zero therefore net work done should also be zero.
I think that's probably a satisfactory answer.
I think that's probably a satisfactory answer.
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