A particle's kinetic energy increases from 8 J to 20 J as it undergoes a displacement r = (3 i - 2 j ) m . The constant force acting on the particle is given by F = (2 i + F_y j ) N . The value of the unknown force component F_y is

Step-by-step solution

According to the work-energy theorem, the net work done on a particle equals its change in kinetic energy: W = K = K_f - K_i Given K_i = 8 J and K_f = 20 J : W = 20 - 8 = 12 J The work done by the constant force is given by the dot product of force and displacement: W = F r W = (2 i + F_y j ) (3 i - 2 j ) W = (2)(3) + (F_y)(-2) = 6 - 2F_y Equating the two expressions for work done: 6 - 2F_y = 12 -2F_y = 6 F_y = -3 N Answer: -3 N