NIMCET 2020 Question Paper with solutions

Question : 33 of 120

Marks: +1, -0

Forces of magnitude 5, 3, 1 units acts in the directions

6i + 2j + 3k, 3i - 2j + 6k, 2i - 3j - 6k

respectively on a particle which is displaced the point (2, -1, -3) to (5, -1,1). The total work done by the force is

Solution:

Given,

A↖{→}=2i↖{⋀}-j↖{⋀}-3k↖{⋀}

B↖{→}=5i↖{⋀}-j↖{⋀}+k↖{⋀}

{AB}↖{→}=3i↖{⋀}+4k↖{⋀}

F↖{→}_1=5/7(6i↖{⋀}+2j↖{⋀}+3k↖{⋀})

F↖{→}_2=3/7(3i↖{⋀}-2j↖{⋀}+6k↖{⋀})

F↖{→}_3=1/7(2i↖{⋀}-3j↖{⋀}-6k↖{⋀})

F↖{→}=F↖{→}_1+F↖{→}_2+F↖{→}_3

{41i↖{⋀}+j↖{⋀}+27k↖{⋀}}/7

F↖{→}.{AB}↖{→}={41i↖{⋀}+j↖{⋀}+27k↖{⋀}}/7(3i↖{⋀}+4k↖{⋀})

{123+108}/7=231/7=33 units

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