A particle of mass 2 kg moves in the horizontal x-y plane under the influence of a net force F = (kx i + 4y j ) N , where k is a constant. The particle moves from the initial position (0, 1) m to the final position (2, 2) m , and its speed increases from 2 m/s to 4 m/s . The value of the constant k is :

Step-by-step solution

Using the Work-Energy Theorem, the net work done on the particle equals its change in kinetic energy. The change in kinetic energy is: KE = 1 2 m(v_f^2 - v_i^2) KE = 1 2 (2)(4^2 - 2^2) = 16 - 4 = 12 J The work done by the variable force is the line integral from (0, 1) to (2, 2) : W = F d r = ₀² kx , dx + ₁² 4y , dy W = [ kx^2 2 ]₀² + [ 4y^2 2 ]₁² W = k 2 (4 - 0) + 2(4 - 1) W = 2k + 6 Equating the work done to the change in kinetic energy: 2k + 6 = 12 2k = 6 k = 3 Answer: 3