A bead of mass 1 kg is threaded on a smooth rigid wire shaped as a parabola y = x^2 in the horizontal x-y plane. It is pulled by a non-conservative force F = (3y i + 6x j ) N , where x and y are in meters. If the bead starts from rest at the origin (0,0) , its kinetic energy when it reaches the position (2,4) m is :

Step-by-step solution

The normal force from the smooth wire is perpendicular to the displacement and does zero work. By the Work-Energy Theorem, the kinetic energy of the bead equals the work done by the applied force. The work done is given by the line integral: W = F d r = (3y , dx + 6x , dy) The bead is constrained to move along the path y = x^2 . Differentiating this gives dy = 2x , dx . Substituting y and dy into the integral to express everything in terms of x : W = ₀² ( 3(x^2) , dx + 6x(2x , dx) ) W = ₀² (3x^2 + 12x^2) , dx W = ₀