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Question

A particle acted upon by constant forces F1=4i^+j^-3k^ and F2=3i^+j^-k^ is displaced from the point r1=i^+2j^-3k^ to point r2=5i^+4j^-k^. The total work done by the forces in SI unit is


A

20

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B

24

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C

50

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D

30

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E

35

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Solution

The correct option is B

24


Step 1: Given data.

Force, F1=4i^+j^-3k^

Force, F2=3i^+j^-k^

Displacement, r1=i^+2j^-3k^

Displacement, r2=5i^+4j^-k^

Step 2: Find the total work done by the forces.

The net force is,

F=F1+F2F=4i^+j^-3k^+3i^+j^-k^F=7i^+2j^-4k^

Net Displacement is,

d=r2-r1d=(5i^+4j^-k^)-(i^+2j^-3k^)d=4i^+2j^+2k^

The work done is given by,

W=F.dW=(7i^+2j^-4k^).(4i^+2j^+2k^)W=28+4-8W=24

Hence, option B is correct.


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