Given:
Force applied by Vijay, F1=22 N
Force applied by Gopal, F2=10 N.
Displacement, s=6.5 m
Since, both the forces are acting in the opposite directions, net force is given by:
F=F1−F2
F=22−10=12 N
The work done on the block is given by the product of displacement and the net force applied on it. Hence it is given by:
W=F×s
W=12×6.5=78 J