Question

A force of 14 N acts on a particle along the vector (3i +2j -6k) . If the particle displaces from (0, 0, 0) to (2, 4, 2), the work done (W = F ×S) by force on the particle is

Answer 1

The work done by the force on the particle is 4 Joule

Explanation:

Given

Force of 14 N acts along position vector$\vec r=3\hat i+2\hat j-6\hat k$

Magnitude of the force $|\vec F|$ = 14

Unit vector along $\vec r$

$\hat r=\frac{3\hat i+2\hat j-6\hat k}{\sqrt{3^2+2^2+(-6)^2}}$

$\implies \hat r=\frac{1}{\sqrt{49}}(3\hat i+2\hat j-6\hat k)$

$\implies \hat r=\frac{1}{7}(3\hat i+2\hat j-6\hat k)$

Thus, the force vector will be

$\vec F=|\vec F|\hat r$

$\implies \vec F=14\times \frac{1}{7}(3\hat i+2\hat j-6\hat k)$

$\implies \vec F=2(3\hat i+2\hat j-6\hat k)$

Displacement of particle is from (0,0,0) to (2,4,2)

Therefore, the displacement vector

$\vec d=2\hat i+4\hat j+2\hat k-0$

$\implies \vec d=2(\hat i+2\hat j+\hat k)$

We know that work done

$W=\vec F.\vec d$         (Note: it is a scalar product i.e. work is a scalar quantity)

Therefore,

$W=2(3\hat i+2\hat j-6\hat k).[2(\hat i+2\hat j+\hat k)]$

$\implies W=4(3\times 1+2\times 2-6\times 1)$

$\implies W=4\times (3+4-6)$

$\implies W=4\times 1$

$\implies W=4$ Joule

Hope this answer is helpful.

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Answer 2

Explanation:

IF IN THE QUESTION, THE DISPLACEMENT COORDINATES ARE GIVEN (0,0,0) AND (2,4,-2),

THEN THE WORK DONE (W) = F . S = 52J

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