JEE Main Physics question
A position-dependent force F = 3x² − 2x + 7 acts on a body of mass 7 kg and displaces it from x = 0 m to x = 5 m. The work done on the body is x′ joule. If both F and x are measured in SI units, the value of x′ is
Answer: (A)
Solution
W = ∫₀⁵ (3x²−2x+7) dx = [x³−x²+7x]₀⁵ = 125 − 25 + 35 = 135.