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Question

A particle moves from a point (2^i+5^j) m to (4^j+3^k) m when a force of (4^i+3^j) N is applied. Then, work done by the force is

A
8 J
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B
11 J
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C
5 J
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D
2 J
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Solution

The correct option is C 5 J
Initial position vector, r1=(2^i+5^j) m
Final position vector, r2=(4^j+3^k) m
Displacement of the particle is,
r=r2r1=(2^i1^j+3^k) m
F=(4^i+3^j) N
Since the force is constant, the work done by the force is given by,
W=F.r
W=(4^i+3^j).(2^i1^j+3^k)
W=(4×2)+(3×1)+(0×3)
W=5 J
Work done by the force is 5 J

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