A man drags a 25.0kg crate, by a rope, across the floor whose coefficient of kinetic friction is 0.250. The tension in the rope is 200.0 N at an angle of 30.0° to the vertical. What is the net work done on the box as it travels 10.0 m? well I broke the tension force into the x and y components with, these equations. cos60x200 = 100N sin60x200=173.205N also the Frictional force in the opposite direction of +x movement is fk=ukFN fk= 0.25 x 25kg x 9.8m/s^2 = 61.25N This is where I'm stuck, I thought I could subtract the x components and create a new triangle, find the new angle and determine the work done with W= Force x Distance cos theta. The answer I'm getting seems to be wrong. Please answer this as soon as possible, I have a midterm tommorow and I'm struggling to teach myself. Thank you.