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Q.

A force  F →=6i^+2j^−3k^ acts on a particle and produces a displacement of s →=2i^−3j^+xk^.  If the work done is zero, the value of x is

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a

– 2

b

1/2

c

6

d

2

answer is D.

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Detailed Solution

W=F→.s→=(6i^+2j^−3k^).(2i^−3j^+xk^)=012−6−3x=0 ⇒x=2
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