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Question

# A particle moves from a point (2^i+^k) m to (4^i+3^j−^k) m when a force of (3^i+^j) N is applied. The 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
9 J
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Solution

## The correct option is D 9 J Since the force is constant here, work done by the force will be given by W=→F.→d ...(i) where displacement is given as →d=→r2−→r1 & →ri represents the position vector of particle. Given, →F=3^i+^j N →r2=4^i+3^j−^k m & →r1=2^i+^k m →d=→r2−→r1 =2^i+3^j−2^k m From Eq.(i) we get, W=(3^i+^j).(2^i+3^j−2^k) ⇒W=(3×2)+(1×3)+(0×(−2)) ∴W=9 J Work done by the force is 9 J.

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