NIMCET 2014 Question Paper with solutions

Question : 5 of 120

Marks: +1, -0

Constant forces P =

2i↖{\^} - 5j↖{\^} + 6k↖{\^}

and Q =

-i↖{\^} + 2j↖{\^} - k↖{\^}

act on a particle. The work done when the particle is displaced from A whose position vector is

4i↖{\^} - 3j↖{\^} - 2k↖{\^}

to B whose position vector is

6i↖{\^} + j↖{\^} - 3k↖{\^}

, is

Solution:

We have,
P =

2i↖{\^} - 5j↖{\^} + 6k↖{\^}

Q =

-i↖{\^} + 2j↖{\^} - k↖{\^}

Position vector of
A =

4i↖{\^} - 3j↖{\^} - 2k↖{\^}

and B =

6i↖{\^} + j↖{\^} - 3k↖{\^}

d = AB =

2i↖{\^}+4j↖{\^}-k↖{\^}

∴ F = P + Q =

i↖{\^}-3j↖{\^}+5k↖{\^}

Work done = F-d
=

(i↖{\^}-3j↖{\^}+5k↖{\^}).(2i↖{\^}+4j↖{\^}-k↖{\^})

= 2 - 12 - 5 = 15 units

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