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JEE Main 2022 · 24 Jun · Shift 1, Physics question 39

A particle experiences a variable force F⃗=(4xi^+3y2j^)\vec F=\left(4x\hat i+3y^2\hat j\right) in a horizontal xx-yy plane. Assume distance in meters and force is newton. If the particle moves from point (1,2)(1,2) to point (2,3)(2,3) in the xx-yy plane, then Kinetic Energy changes by :

Choose one option
Answer and worked solution

Official answer: C

Work-energy theorem. ΔK=W=∫Fx dx+∫Fy dy\Delta K=W=\int F_x\,dx+\int F_y\,dy (the force is conservative, so the path does not matter).

Integrate. ∫124x dx=2(4−1)=6\int_1^2 4x\,dx=2(4-1)=6 J and ∫233y2 dy=27−8=19\int_2^3 3y^2\,dy=27-8=19 J.

Total. ΔK=25\Delta K=25 J.

(C)

Topics:Work by constant and variable forcesWork-energy theorem
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