# An object is displaced from position vector r1 = (2i + 3j)m to r2 = (4i + 6j)m under a force F = (3x2i + 2yj)N. Find the work down by this force.

Sujit Kumar
111 Points
4 years ago
NOTE : '[' is the integration symbol

W.D = 2[4 3xdx + 3[6 2y dy
= x2|+ y2 3|= 64 – 8 + 36 – 9 = 56 + 27 = 83J
Khimraj
3007 Points
4 years ago
Work done = $\int_{x1}^{x2}F_{x}dx + \int_{y1}^{y2}F_{y}dy$ = $\int_{2}^{4}3x^{2}dx + \int_{3}^{6}2ydy$ = 83J
Hope it clears.