Practice question
Question
A force F\=4i^+6j^N acts on a particle moving along d\=−3i^+2j^m. What is the work done?
Explanation
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 0 J. Work-energy theorem W=ΔK confirms. So option B is correct.