A particle of mass 1 kg , initially at rest at the origin, is acted upon by a two-dimensional variable force given by F = (2x i + 3y^2 j ) N , where x and y are in metres. The speed of the particle when it reaches the coordinates (2 m , 2 m ) is:

Step-by-step solution

The work done by a two-dimensional variable force is calculated by integrating the force components along their respective axes. W = F d r = (F_x dx + F_y dy) Given F = (2x i + 3y^2 j ) N , the work done from (0,0) to (2,2) is: W_x = ₀² 2x dx = [ x^2 ]₀² = 4 J W_y = ₀² 3y^2 dy = [ y^3 ]₀² = 8 J Total work done, W = W_x + W_y = 4 + 8 = 12 J According to the work-energy theorem, the total work done is equal to the change in kinetic energy: W = K = 1 2 m v^2 - 1 2 m u^2 Since the particle starts from rest ( u = 0 ) an