A person pushes a 13.0 kg lawn mower at constant speed with a force of 84.0 N directed along the handle, which is at an angle of = 49.0° to the horizontal. (a) Calculate the horizontal retarding force on the mower (b) Calculate the normal force exerted vertically upward on the mower by the ground. (c) Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.0 seconds (assuming the same retarding force). Here's what I have so far... a) FPx = 84 cos 49 = 55.11 N FPy = 84 sin 49 = 63.396 N ∑ F = ma N – mg = ma N = mg = 13.0(9.8) = 127.4 N b) Ffr = (coefficient of friction)FN = (0.30)(127.4) = 38.22 N c) I have no idea where to even start for this question! Please help!!!