Basically my instructor is confusing the heck out of me because I highly dislike how this course is structured. I dislike how in this specific chapter we are not taught: [tex]\Sigma[/tex]W = [tex]\Delta[/tex]Energy However, using this equation and my previous knowledge I can't seem to get the right answer in the following problem unless I assume there is no [tex]\Delta[/tex]PE which isn't true since [tex]\Delta[/tex]height = .150 When using the relationship you see below, do we assume that the Work done by gravity doesn't exist since we compensate for it under the change in potential energy? 1. The problem statement, all variables and given/known data In Figure 7-32, a constant force of magnitude 82.0 N is applied to a 3.00 kg shoe box at angle f = 53.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by when the box has moved through vertical distance h = 0.150 m?? 2. Relevant equations http://www.lowellphysics.org/beta/T...rk/Problems/c07x7_11.xform_files/nw0314-n.gif [tex]\Sigma[/tex]W = [tex]\Delta[/tex]Energy [tex]\Sigma[/tex]W = [tex]\Delta[/tex]KE + [tex]\Delta[/tex]PE Wg + Wapp = 0(constant speed) + [tex]\Delta[/tex]PE Wg + Wapp = [tex]\Delta[/tex]PE 3. The attempt at a solution Using the above formula: Wg + Wapp = [tex]\Delta[/tex]PE Wg + Wapp = (3)(g)([tex]\Delta[/tex]height) Wg + Wapp = (3)(g)(-.15) Wg + Wapp = -4.4145 Wg = F * d * cos[tex]\Phi[/tex] Wg = mg * (.15) * cos(0) Wg = (3)(9.81) * (.15) Wg = 4.4145 Wapp = F * d * cos[tex]\Phi[/tex] Wapp = 82cos53 * (.15) * cos0 Wapp = 7.4 This relationship doesn't make any sense... 4.41 + 7.4 [tex]\neq[/tex] -4.4145 Basically, if I assume that there is no change in potential energy, however, the answer does end up coming out right... kinda of.