Work done by a constant force is defined by W=F(delta d)(cos theta), where F is the magnitude of force and delta d is the magnitude of the displacement (not distance). (I got the definition from a textbook) Now consider the work done by friction for, 1) a box sliding across the floor to the right by 5m and and slides back to the original position The magnitude of the displacement is 0, so the work done by friction is zero? This doesn't seem to be true...how come I got the wrong answer? Now consider the work done by gravity for, 2) a rocket near the surface of the earth travels in a straight line 40 degrees above the horizontal for 1000m When using W=F(delta d)(cos theta) to find the work done by gravity, should I substitute the magnitude of the displacement (1000m) or just the magnitude of the vertical component of the displacement (1000sin130 m)? Can someone clarify these concepts? Thanks!

You forgot about the 'cos(theta)' term. I prefer W=F<sup>.</sup>d, (the dot product of the force vector and displacement vector) over your textbook equation. The two equations are identical, but the textbook equation hides the dot product in the cos(theta) term. In your first problem, the magnitude of the friction force is indeed constant as the block slides back and forth. Is the same true of the force vector? What is the angle between the force and displacement vectors in your second problem?

1) But wouldn't the magnitude of the displacement be zero? 2) Sorry...I mean 130 degrees But which one should I substitute in for the magnitude of the displacement, 1000 m or 1000sin130 m?

1) Your equation (and mine; they are the same thing) are only valid when the force vector is constant. Is the force vector constant in this problem? If the answer is no, try breaking the problem into parts where the answer is 'yes' for each part. 2) This is much easier visualized as a vector problem. Draw the vectors. You do know that a<sup>.</sup>b = |a| |b| cos(theta), where theta is angle between the two vectors?

1) I see what you meant, thanks! 2) Fg points straight down, and the motion is 40 degrees above horizontal, so 90+40=130 degrees is the angle between F and delta d, but what should I substitute in for delta d?