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Physics Question on work, energy and power

A particle acted upon by constant forces (4i^+j^−3k^)( 4\hat{i}+\hat{j}-3\hat{k} ) and (3i^+j^−k^)( 3\hat{i}+\hat{j}-\hat{k} ) is displaced from the point (i^+2j^+3k^)( \hat{i}+2\hat{j}+3\hat{k} ) to the, point (5i^+4j^+k^) (5\hat{i}+4\hat{j}+\hat{k} ) . The total work done by the forces in SISI unit is

A

20

B

40

C

50

D

30

Answer

40

Explanation

Solution

Here, F1⃗=4i^+j^−3k^,F2⃗=3i^+j^−k^\vec{F_{1}} = 4\hat{i}+\hat{j}-3\hat{k}, \vec{F_{2}} = 3\hat{i}+\hat{j}-\hat{k}
r1⃗=i^+2j^+3k^\vec{r_{1}} = \hat{i}+2\hat{j}+3\hat{k}, r2⃗=5i^+4j^+k^\vec{r_{2}} = 5\hat{i}+4\hat{j}+\hat{k}
Displacement, r⃗=r2⃗−r1⃗\vec{r} = \vec{r_{2}}-\vec{r_{1}}
=(5i^+5j^+k^)−(i^+2j^+3k^)= \left(5\hat{i}+5\hat{j}+\hat{k}\right)-\left(\hat{i}+2\hat{j}+3\hat{k}\right)
=4i^+2j^−2k^= 4\hat{i}+2\hat{j}-2\hat{k}
Work done by the forces,
W=[F1⃗+F2⃗]⋅r⃗W = \left[\vec{F_{1}}+\vec{F_{2}}\right]\cdot\vec{r}
=[(4i^+j^−3k^)(3i^+j^−k^)]⋅(4i^+2j^−2k^)= \left[\left(4\hat{i}+\hat{j}-3\hat{k}\right)\left(3\hat{i}+\hat{j}-\hat{k}\right)\right]\cdot\left(4\hat{i}+2\hat{j}-2\hat{k}\right)
=(7i^+2j^−4k^)⋅(4i^+2j^−2k^)= \left(7\hat{i}+2\hat{j}-4\hat{k}\right)\cdot\left(4\hat{i}+2\hat{j}-2\hat{k}\right)
=28+4+8=40 J= 28+4+8 = 40\,J