[SOLVED] Mininum Work to Pull Crate 1. The problem statement, all variables and given/known data A student could either pull or push, at an angle of 30 degrees from the horizontal, a 50 kilo crate on a horizontal surface, where the coefficient of kinetic friction between the crate and the surface is 0.2. The crate is to be moved a horizontal distance of 15m. (a) Compared with pushing, pulling requires the student to do (1) less, (2) the same, or (3) more work. (b) Calculate the minimum work required for both pulling and pushing. 2. Relevant equations 3. The attempt at a solution This is a problem that has been posted before but remained unsolved. I just need assistance with part b. For part a, I know that pulling requires less work because pushing creates a greater normal force and therefore, friction. Onto part B: Let: uk = coefficient of kinetic friction, .20 m = mass of crate, 50kg g = acceleration constant, 9.8m/s^2 fk = force of kinetic friction N = normal force exerted by the crate F = unknown minimum force Calculate the minimum work required for pulling: 1. The minimum work occurs when the horizontal force equals the force of kinetic friction. The force of kinetic friction is: fk = uk*N = uk*(mg-F*sin(30)) Setting fk equal to the horizontal component of the applied force: uk*(mg - F*sin(30) ) = F*cos(30) ==> F = 101 N The minimum work in this case should be 101 N*15 m = 1521 J. But it turns out that this wasn't correct. What am I doing wrong?