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 joule is :

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Solution

The correct option is B + $7$
$W = ¯ F . ¯ x = (5^i + 3^j + 2^k) . ((2 - 0)^i - (1 - 0)^j) = 10 - 3 = 7 j o u l e s$

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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 joule is :

The correct option is B +7W=¯F.¯x=(5^i+3^j+2^k).((2−0)^i−(1−0)^j)=10−3=7joules

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 joule is :

The correct option is B + $7$
$W = ¯ F . ¯ x = (5^i + 3^j + 2^k) . ((2 - 0)^i - (1 - 0)^j) = 10 - 3 = 7 j o u l e s$