JEE Main Physics question
A position dependent force F = 7 − 2x + 3x² newton acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joules is
Answer: (D)
Solution
W = ∫₀⁵ (7 − 2x + 3x²) dx = [7x − x² + x³]₀⁵ = 35 − 25 + 125 = 135 J.