The net force acting on the block is the vector sum of the two forces:
F=F1+F2=(2i^+3j^+4k^)+(3i^−j^−2k^)=5i^+2j^+2k^ N
The block is moved a distance of 25 m along the direction of the vector 3i^−4j^. The unit vector along this direction is:
n^=32+(−4)23i^−4j^=53i^−4j^
The displacement vector s is the magnitude multiplied by the unit vector:
s=25n^=25(53i^−4j^)=15i^−20j^ m
The work done W is the dot product of the net force and the displacement vector:
W=F⋅s=(5i^+2j^+2k^)⋅(15i^−20j^)
W=(5×15)+(2×−20)+(2×0)
W=75−40=35 J
Answer: 35