1. The problem statement, all variables and given/known data A woman pushes a 38kg box on a horizontal floor with a force F⃗ that is directed 30° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.53. If the initial speed of the box is 7.89m/s and the box accelerates at 1.63m/s^2 , what is the net work done on the box by all forces acting on it after it has traveled a distance of 2.51m ? The answer provided is 155.47J. 2. Relevant equations F=ma? Fk = ukN W= fdcos(x) W= fd W= Kf - Ki 3. The attempt at a solution Method 1: F=ma Fcos(30) - Fk = 38 (1.63) Fcos(30) - 0.53N= 38 (1.63) Finding the Normal Force: N-mg-Fsin(30) =0 N =38 (9.81) + Fsin(30) Subbing N back into the equation Fcos(30) - 0.53[38 (9.81) + Fsin(30)]= 38 (1.63) F = 38(1.63) + 0.53(9.81)(38) / cos(30) - 0.53sin(30) Work: Fcos(30) x 2.51 = 1541.18J Method 2: W= Kf - Ki = 0.5(38)(54.07) - 0.5(38)(62.25) = -155.47J Can somebody tell me why I got the wrong answer using the first method, yet when I use the second method I get the answer with a minus sign (which indicates negative work?) ? If that is the case, then shouldn't the answer then be -155.47J? Thank You!