Question

Find the work done in moving a particle along a vector S =4i_j +7k m. If the applied force is vector F=i+2j_k

Answer 1

F = i + 2j - k
S = 4i - j + 7k

Work = F · S ………[It is dot product]
= (i + 2j - k) · (4i - j + 7k)
= 4 - 2 - 7
= -5 joules

Work done is negative and it’s magnitude is 5 joules.

Answer 2

Answer:

The work done in moving a particle is -5J.

Explanation:

Given the displacement of a particle, $\vec S =4\hati-\hat j +7\hat k ~m$

The force applied on the particle, $\vec F=\hat i+2\hat j-\hat k ~N$

Work done is defined as the displacement of the particle in the direction of applied force.

Mathematically written as $W=\vec F . \vec S$

$=(\hat i+2\hat j-\hat k) .(4\hati-\hat j +7\hat k)$

$=1\times 4+2\times (-1) +(-1) \times 7$  

$= 4-2 - 7=-5~N.m$

Therefore, the work done in moving a particle is -5N.m or 5J.