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Question

An object is displaced from position r1=2i^+3j^m to r2=4i^+6j^m under a force F=3x2i^+2yj^N then work done by the force is


A

24J

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B

33J

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C

83J

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D

45J

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Solution

The correct option is C

83J


The explanation for correct option

Step 1: Given

  1. The object is displaced from position r1=2i^+3j^m to r2=4i^+6j^m
  2. Applied force F=3x2i^+2yj^N

Step 2: Formula used

W=F·dr=F·dxi^+dyj^

Where W is the work done and F is the applied force

Step 3: Find the work done

The object is displaced from position r1=2i^+3j^m to r2=4i^+6j^m

Therefore x1=2,y1=3,x1=4,y1=6

W=R1R2F·dr=R1R23x2i^+2yj^·dxi^+dyj^=243x2dx+362ydy=x324+y236=64-8+36-9=83J

Hence option C is correct


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