# Work done by the gravity force

I have this question:

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!

## Answers and Replies

Gravity's work is -mgsin30d in this case.
In incline problems gravity has 2 components: 1) y-component, mgcos30, which doesn't do work (perpendicular), and the other one is mgsin30. W=F.d (dot prodoct) or W=Fdcos
(theta). ==> work of this force is mgsin30d(cos180), which is -mgdsin30..

Hi, thanks for your help, but when I do -mgdsin(30), I get -(28 lb)(32 ft/s^2)(15 ft)sin(30) = -6,720 ft-lb, not the -210 ft-lb that the book says I should get. Any ideas?

I have no other idea..b/c i'm sure my anwer is correct :)...maybe the book's answer is wrong or the book's given numbers are not the same as you've posted...
or maybe it wants the answer in jouls not ft*lb...

Notice that the work of gravity is actually mgh, which "h" is the height the object has pulled up, so h=dsin(theta) and the total work equation is the one I posted earlier: -mgdsin30

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