Suppose the same 65.0-kg player instead slides with the same 6.00 m/s initial speed up a 5.00 degree incline, still against a 450 N friction force. Compared with sliding the 2.60 m distance found on level ground, does he slide farther or a shorter distance up the incline, and why?

1. Going uphill, both friction and the gravitational potential energy gain work against the player's initial kinetic energy, since one half m vi squared equals fd plus mgd sin(theta). 2. Because two effects now remove kinetic energy instead of just friction alone, the stopping distance d comes out smaller than the 2.60 m found on level ground. 3. The player therefore slides a shorter distance up the incline, not farther and not the same distance as on level ground. 4. The problem gives the friction force directly as 450 N, so the coefficient of friction is not needed to determine that the distance decreases. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.5 Nonconservative Forces_