Question

A particle moves from a point →r1=(2 m) →i+(3 m) →j→r1=(2 m) →i+(3 m) →j to another point →r2= (3 m) →i+(2 m) →j→r2= (3 m) →i+(2 m) →j acts on it. Find the work done by the force on the particle during the displacement.

Answer 1

The work done by the force on the particle during the displacement is zero.

Explanation:

Step 1:

The vector of initial position is shown by the equation, $r 1 \rightarrow=3 j \rightarrow+2 i \rightarrow=\rightarrow r 1 \rightarrow=2 i \rightarrow+3 j \rightarrow$

The vector of the final position is shown by the equation, $r \rightarrow 2=2 j \rightarrow+3 i \rightarrow r \rightarrow 2$

$=2 j \rightarrow+3 i \rightarrow$

Step 2:

So, the displacement vector,

$r \rightarrow=-r \rightarrow 1+r \rightarrow 2=-(2 i \rightarrow+j \rightarrow)+(3 i \rightarrow+2 j \rightarrow)$

$=i \rightarrow-j \rightarrow$

Step 3:

On the particle, the force acted is $F \rightarrow=5 j \rightarrow+5 i \rightarrow$

So, work done $=F \rightarrow \cdot S \rightarrow=5(-1)+5 \times 1=0$

So, the word done is zero.