Question

A force F=(2i^+5j^+k)N is acting on a particle.the particle is first displaced from(0,0,0) to(2,2,0)m along the path x=y and then from (2,2,0)m to (2,2,2)m along the path x=2m,y=2m. The total work done in the complete path is

Answer 1

Answer:

applied the formulae work done =force multiply by distance

Answer 2

Answer:

16

Explanation:

we know work equals to force × displacement.

net displacement is (2 i + 2 j + 2 k)

now F= 2 i + 5 j + k

hence work equals to 4 + 10 + 2=16J