A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force. a. when the ramp is frictionless. b) when the coefficient of kinetic friction is 0.18. Fnet = ma Fg= mg Ff=uk(Fn) a) With upwards and right as positive and downwards and left as negative i) After drawing a diagram, I broke the Fnet into its x and y components. I determined Ffrictionx= Ffriction(cos28), Ffrictiony= Ffriction(sin28). I will not include these in the horizontal and vertical components as it is a frictionless surface. Fnet y = ma Fnormal - Fg = ma Fn= m(a-g) The object is not accelerating in the horizontal direction so a=0 Fn= m (-g) Fn = 90kg (-(-9.8m/s^2))= 882N. So the vertical force is 882N in the upward direction. Fnet x= ma Fpush= ma Fp=(90kg)(a). I don't know what the acceleration is in the horizontal direction so I'm not sure where to go from here. b). Again, break into horizontal and vertical components with up/right as positive, down/left as negative. Fnet y = ma Fpush - Ffrictionx = ma Fp - Ff(Cos28) = ma Fp - (cos28) (uk)(Fn)= ma Fp - (cos28(uk)(-mg))= ma Fp= ma +140.2 ... I'm not sure if I'm on the right track here, or if I am, where to go from here.