Three constant forces`vecF_(1)=2hati-3hatj+2hatk,vecF_(2)=hati+hatj-hatk` and j - 2k in newton displace a particle from (1,-1,2) to (-1,-1, 3) to (2,2,0) (displacement being in metres). The total work done by the forces is, if displacement is along straight path:

Options

A. 2 J
B. 3 J
C. 4 J
D. 5 J

Answer

d

Explanation

Here net resultant force `vecF=vecF_1+vecF_2+vecF_3=(6hati-hatj-hatk)N`
Net displacement
=`(2-1)hati+(2+1)hatj+(0-2)hatk=hati+3hatj-2hatk`
Now work done W
=`(6hati-hatj-hatk).(hati+3hatj-2hatk)=(6-3+2)=5J`