Question

A body is displaced from, A=(2m, 4m, -6 m) to r_(B)=(6hati-4hatj + 2hatk)m under a constant force, F=(2hati + 3hatj-hatk)N. Find the work done.

Answer 1

The work done is:

• Initial position vector = (2 i + 4 j - 6 k) m

       Final position vector = (6 i - 4 j +2 k) m

• Displacement, s = Final Position Vector - Initial Position Vector

                                   = (4 i - 8 j +8 k) m

• Given, Force, F = (2 i +3 j - k) N

• Work done = F.s = (2 i +3 j - k) N . (4 i - 8 j +8 k) m = (2*4)-(3*8)-(1*8) = -24 J