Aforce of 10 n is applied horizontally to a 2.0-kg mass on a level surface. the coefficient of kinetic friction between the mass and the surface is 0.20. if the mass is moved a distance of 10 m, what is the change in its kinetic energy?

Force = 10 N; Mass = 2 kg; Coefficient of kinetic friction = 0.20; Distance s = 10 m Force m x g = 2 x 9.8 = 19.6 Multiplying force with Coefficient of friction that gives force required = 19.6 x 0.2 = 3.92 N Resultant force = 10 - 3.92 = 6.08 Acceleration a = F / m = 6.08 / 2 = 3.04 We have equation v^2=u^2+2as initial velocity will be 0 thatis u^2 = 0, so v^2 = 2as = 2 x 3.04 x 10 = 60.8 Final Kinetic energy KEf = 1/2 x m x v^2 = 1/2 x 2 x 60.8 = 60.8 J Initial Kinetic energy KEi = 1/2 x m x u^2 = 1/2 x 2 x 0 = 0 So change in kinetic energy = 60.8 J

The answer is orbit around nucleus