A 1 kg block subjected to two simultaneous forces (2i^+3j^+4k^) N and (3i^-j^-2k^) N is moved a distance of 25 m along (3i^-4j^) direction. The work done in this process is _______ J. [2026]
(35)
F→1=(2i^+3j^+4k^) NF→2=(3i^-j^-2k^) N}⇒F→net=(5i^+2j^+2k^) N
(distance) d=25 m along (3i^-4j^) direction
∴ displacement d→=25(3i^-4j^9+16)=25(3i^-4j^5) m
d→=(15i^-20j^) m
So, work done W=F→net·d→
W=(5i^+2j^+2k^)·(15i^-20j^)=35 J