I must be missing something pretty basic here since I solved a similar problem earlier in the chapter with no problems. The Question: to push a 25.0kg crate up a frictionless incline, angled at 25.0 degrees to the horizontal, a worker exerts a force of 209N, parallel to the incline. As the crate slides 1.5m, how much work is done on the crate by (a) the workers applied force (b) the weight of the crate (c) the normal force (d) what is the total work done on the crate? (a) is the one that is giving me problems. The book lists the answer as 314J The change in d is .63 m given by 1.5 sin 25 The work done by gravity is m*g*d=-154 J This is where I'm running into problems, shouldn't the negative work done by gravity=the work done by the worker? A similar problem earlier in the chapter had a crate being pulled up the ramp by a rope, I don't see how that problem is any different then this one other then the initial info given. I applied the same technique to solving both problems and came up with the right answer for one and the wrong answer for the other. Thanks for any help.