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A 28 lb suitcase is pulled up a frictionless plane inclined at 30 degrees to the horizontal by a force P with a magnitude of 18 lb and acting parallel to the plane. If the suitcase travels 15 ft along the incline, calculate:

(a) the work done on the suitcase by the force P,

(b) the work done by the gravity force,

(c) the work done by the normal force,

(d) the total work done on the suitcase.

I got (a) from W=Pscos(theta), so (18 lb)(15 ft)(cos 0) = 270 ft-lb and part (c) b/c W=0 for the normal force since it's perpendicular to the motion, but I can't get part (b). The answer says it's -210 ft-lb, but I can't seem to get this. I've drawn components, so there's the mg going straight down from the suitcase on the incline, and it's y-component is mgsin(theta) and the x-component is mgcos(theta), but I don't really know how to get the work done by gravity. Any help would be much appreciated. Thanks!