Question

The forces with magnitudes 5,3 and 1 unit are acting on a particle in the direction of the vectors 6i^+2j^+3k^;3i^−2j^+6k^ and 2i−3j^−6k^, respectively. They remain constant while the particle is displaced from the point A(2,−1,−3) to B(5,−1,1). The work done is

• A. (A) 11 units
• B. (B) 33 units
• C. (C) 10 units
• D. (D) 30 units

Answer 1

Answer: (B)

(B) 33 units

Answer 2

Explanation:
∵F→1=5(6i^+2j^+3k^)7F→2=3(3i^−2j^+6k^)7 and F→3=1(2i^−3j^−6k^)7∴F→=F→1+F→2+F→3+2i^−3j^−6k^)=17(30i^+10j^+15k^+9i^−6j^+18k^=17[41i^+j^+27k^] and AB→=5i^−j^+k^−2i^+j^+3k^=3i^+4k^∴ Work done −17[41i^+j^+27k^]⋅[3i^+4k^]−17[123+108]=33 units