Question

A force of magnitude 10 N is acting on a particle along i +j - k. The particle displaces from
A(1,2,3)m to B(4,5,6)m. The work done by force on the particle is
Please help !!​

Answer 1

the answer is 30joules

Answer 2

The work done by the force on the particle is 10√3 J

Explanation:

Given

magnitude of force = 10 N

The force acts along the vector $\hat i+\hat j-\hat k$

Unit vector along this vector

$=\frac{\hat i+\hat j-\hat k}{\sqrt{1^2+1^2+(-1)^2}}$

$=\frac{\hat i+\hat j-\hat k}{\sqrt{3}}$

Therefore, the force vector will be

$\vec F=10\times \frac{\hat i+\hat j-\hat k}{\sqrt{3}}$

$=\frac{10}{\sqrt{3}}(\hat i+\hat j-\hat k)$

Position vector of A = $\hat i+2\hat j+3\hat k$

Position vector of B = $4\hat i+5\hat j+6\hat k$

The displacement vector from A to B

= Position vector of B -  Position vector of A

or $\vec {ds}=(4\hat i+5\hat j+6\hat k)-(\hat i+2\hat j+3\hat k)$

$\implies \vec{ds}= 3\hat i+3\hat j+3\hat k$

$\implies \vec{ds}=3(\hat i+\hat j+\hat k)$

Therefore, the work done

$W=\vec F.\vec {ds}$

$\implies W=\frac{10}{\sqrt{3}}(\hat i+\hat j-\hat k).3(\hat i+\hat j+\hat k)$

$\implies W=\frac{10\times 3}{\sqrt{3}}\times (1+1-1)$ Joule

$\implies W=10\sqrt{3}$ Joule

Hope this answer is helpful.

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