Question

a force F=(5i-3j+2k)N moves a particle from r1=(2i+7j+4k)m to r2=(5i+2j+8k)m.The work done by the force is,

Answer 1

Answer:

Given

F= 5i-3j+2k

now, distance(x) =r2-r1

X= (5i+2j+8k)-(2i+7j+4k)

=3i-5j+4k

Again,

W= F.X

= (5i-3j+2k).(3i-5j+4k)

= 38 joule