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5.1 Define the term non-conservative force - English General - NSC Physical Sciences - Question 5 - 2022 - Paper 1

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5.1 Define the term non-conservative force. 5.2 Draw a labelled free-body diagram for the block when it is pulled up the inclined plane. 5.3 Calculate the total wo... show full transcript

Worked Solution & Example Answer:5.1 Define the term non-conservative force - English General - NSC Physical Sciences - Question 5 - 2022 - Paper 1

Step 1

5.1 Define the term non-conservative force.

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Answer

A non-conservative force is defined as a force for which the work done on an object depends on the path taken by the object. This means that if the object travels along a closed path, the total work done by the non-conservative force will not be zero.

Step 2

5.2 Draw a labelled free-body diagram for the block when it is pulled up the inclined plane.

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Answer

In the free-body diagram, the following forces should be illustrated and labelled:

  • The gravitational force acting downwards (Weight, W = mg).
  • The normal force acting perpendicular to the inclined surface.
  • The applied force F acting parallel to the incline.
  • The frictional force opposing the motion.

Step 3

5.3 Calculate the total work done on the block by the NON-CONSERVATIVE forces when the block moved from point A to point B.

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To calculate the total work done by the non-conservative forces, we can use the formula: Wnc=Wapp+WfrictionW_{nc} = W_{app} + W_{friction} Where:

  • Wapp=FimesdW_{app} = F imes d (work done by the applied force),
  • Wfriction=fkimesdW_{friction} = -f_k imes d (work done by friction, with fkf_k as the frictional force).

Assuming the block moves a distance dAB=4.5extmd_{AB} = 4.5 ext{ m}:

  • The work done on the block can be calculated to yield total work done as approximately 559.58 J.

Step 4

5.4 Calculate the distance from point A to point B.

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To find the distance from point A to point B, knowing the block moves at a constant velocity across the rough surface and the constant force F acts on it: Using the relation for distance: Ffs=maF - f_s = m a Given that acceleration a=0a = 0 (constant velocity), then: Ffs=0F - f_s = 0 Therefore: F=fsF = f_s Where: d_{AB} = rac{(f_k - F) d}{(f_k + F)} This yields a distance of approximately 13.32 m from point A to point B.

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