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Applications of Dot Product

Work done by ...

Question

Work done by a force $\to F = (^i + 2^j + 3^k) N$ action on a particle in displacing it from the point ${\to r}_{1} =^i +^j +^k$ to the point ${\to r}_{1} =^i -^j - 2^k$

- 1 J

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2 J

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- 13 J

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z e r o

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Solution

The correct option is C $- 13 J$
work = force * displacement,

$W = F \times s$

$s = (i - j - 2 k) - (i + j + k) = - 2 j - 3 k$

$F = i + 2 j + 3 k$

$W = (i + 2 j + 3 k) \times (- 2 j - 3 k)$

$∴ W = - 13 J$

Suggest Corrections

Similar questions

\to F = (^i + 2^j + 3^k)

{\to r}_{1} = (^i +^j +^k)

{\to r}_{2} = (^i -^j + 2^k)

\to F = (^i + 2^j + 3^k) N

r_{1} = (^i +^j +^k)

r_{2} = (^i -^j + 2^k)

F = (^i + 2^j + 3^k) N

_{1} =^i +^j +^k

_{2} =^i -^j + 2^k

\to F = (^i + 2^j + 3^k) N

r_{1} = (^i +^j +^k)

r_{2} = (^i -^j + 2^k)

A (1, 2, 3)

B (1, 3, 4)

{\to F}_{1} =

^i + 2^j +^k

{\to F}_{2} =

^i +^j

A

B

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