A constant forces 3-4-5k- acts on a particle at -2-2k- and moves it to a point on z-axis which is 3 units from origin- The work done is

The correct option is B 16 Net force #160; F⃗=3î+4ĵ 5k̂ ..... (i) Net displacement D⃗=3k̂ (î+2ĵ+2k̂) = î 2ĵ+k̂ ..... (ii) Hence th ...

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Standard XII

Physics

Work by Multiple Forces

A constant fo...

Question

A constant forces $3^i + 4^j - 5^k$ acts on a particle at $^i + 2^j + 2^k$ and moves it to a point on $z$ -axis which is $3$ units from origin. The work done is

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Solution

The correct option is B $- 16$
Net force $\to F = 3^i + 4^j - 5^k$ ..... $(i)$
Net displacement
$\to D = 3^k - (^i + 2^j + 2^k)$
$= -^i - 2^j +^k$ ..... $(i i)$
Hence the net work done
$\to F . \to D$ $= (3^i + 4^j - 5^k) . (-^i - 2^j +^k)$
$= - 3 - 8 - 5$
$= - 16$

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A constant forces 3^i+4^j−5^k acts on a particle at ^i+2^j+2^k and moves it to a point on z-axis which is 3 units from origin. The work done is

The correct option is B −16Net force →F=3^i+4^j−5^k.....(i)Net displacement→D=3^k−(^i+2^j+2^k)=−^i−2^j+^k.....(ii)Hence the net work done →F.→D =(3^i+4^j−5^k).(−^i−2^j+^k)=−3−8−5=−16

A constant forces $3^i + 4^j - 5^k$ acts on a particle at $^i + 2^j + 2^k$ and moves it to a point on $z$ -axis which is $3$ units from origin. The work done is

The correct option is B $- 16$
Net force $\to F = 3^i + 4^j - 5^k$ ..... $(i)$
Net displacement
$\to D = 3^k - (^i + 2^j + 2^k)$
$= -^i - 2^j +^k$ ..... $(i i)$
Hence the net work done
$\to F . \to D$ $= (3^i + 4^j - 5^k) . (-^i - 2^j +^k)$
$= - 3 - 8 - 5$
$= - 16$