A force $\to F = (5^i + 3^j + 2^k) N$ is applied over a particle which displaces it from its origin to the point $\to r = (2^i -^j) m$ . The work done on the particle in joules is:

- 7

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+ 7

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+ 10

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+ 13

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Solution

The correct option is B $+ 7$
$Step 1: Workdone as dot product$
$W = \to F . \to Δ r$
$= \to F . (_{f} -_{i})$

$Step 2: Calculation for Workdone$
$W = (5^i + 3^j + 2^k) . ((2^i -^j) - (0^i + 0^j))$
$= (5^i + 3^j + 2^k) . (2^i -^j)$
$= 10 - 3 = + 7 J$
Hence option B is correct.

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A force →F=(5^i+3^j+2^k)N is applied over a particle which displaces it from its origin to the point →r=(2^i−^j)m. The work done on the particle in joules is:

The correct option is B +7Step 1: Workdone as dot product W=→F.→Δr =→F.(→rf−→ri)Step 2: Calculation for Workdone W=(5^i+3^j+2^k).((2^i−^j)−(0^i+0^j)) =(5^i+3^j+2^k).(2^i−^j) =10−3=+7JHence option B is correct.

A force $\to F = (5^i + 3^j + 2^k) N$ is applied over a particle which displaces it from its origin to the point $\to r = (2^i -^j) m$ . The work done on the particle in joules is:

The correct option is B $+ 7$
$Step 1: Workdone as dot product$
$W = \to F . \to Δ r$
$= \to F . (_{f} -_{i})$

$Step 2: Calculation for Workdone$
$W = (5^i + 3^j + 2^k) . ((2^i -^j) - (0^i + 0^j))$
$= (5^i + 3^j + 2^k) . (2^i -^j)$
$= 10 - 3 = + 7 J$
Hence option B is correct.