A 2.00 kg particle moves along +x with speed 1.00 m s⁻¹ at x = 0. From x = 0 to x = 3.00 m the net force is F(x) = 12x − 3x² newtons, with x in metres. Determine the work, final speed, position and value of the maximum force, and the impulse during the motion if the travel time is 1.20 s and the final velocity remains positive. Explain why the impulse cannot be found from the area under F(x) without time information.

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Work = 27.0 J, final speed = 5.29 m s⁻¹, maximum force 12.0 N at x = 2.00 m, and impulse = 8.58 N s. Work is the spatial integral: W = ∫₀³(12x−3x²)dx = [6x²−x³]₀³ = 54−27 = 27.0 J. The initial kinetic energy is ½(2.00)(1.00)² = 1.00 J. Work–energy gives K_f = 28.0 J, so v_f = √(2K_f/m) = 5.29 m s⁻¹. Differentiate the force: dF/dx = 12−6x. The interior maximum occurs at x = 2.00 m and has F = 12(2)−3(2²) = 12.0 N. Impulse equals momentum change, not spatial force area: J = m(v_f−v_i) = 2.00(5.29−1.00) = 8.58 N s. The area under F against x is work because dx is displacement. Impulse requires ∫F dt; the supplied travel time alone would not determine it from an F–x curve, but the momentum result does.

Integrate a position-dependent force

Choose your AP course · School year 2026–27 · May 2027 exam

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