A force F = (5 + 3y²) N acts on a particle in the y-direction. Find the work done as the particle moves from y = 2 m to y = 5 m, in joules.
Physics · Circular Motion
JEE Mains JEE Advanced

Question

A force F = (5 + 3y²) N acts on a particle in the y-direction. Find the work done as the particle moves from y = 2 m to y = 5 m, in joules.

G1 Physics · Accepted Answer

Work done: W = ∫ F dy = ∫₂⁵ (5 + 3y²) dy = [5y + y³]₂⁵ = (25 + 125) − (10 + 8) = 150 − 18 = 132 J Final answer: 132.