a box sliding on a horizontal surface with a speed of 6m/s encounters an incline plane that makes a 15 degree angle with the horizontal line. What is the max distance that will be travelled by the box up the incline before coming to rest if the coefficient of kinetic friction is 0.25? The answer is d=3.67m, however when I try to solve for it I get 4.95m.
Wnc=(delta)KE + (delta)PE
Wfn + Wfk = [(1/2)mvf^2 - (1/2)mvi^2] + [mghf - mghi]
0 + (fk)(d)cos(180) = 0 - (1/2)mvi^2 + mghf - 0
(Uk)(fn)(d)cos(180) = - (1/2)mvi^2 + mghf
-(Uk)(mgcos(15))(d) = - (1/2)mvi^2 + mg(d)(sin(15)
-(Uk)(g)(d)(cos(15)) - (g)(d)(sin(15)) = - (1/2)vi^2
d = (vi^2) / 2 (Uk)(g)(cos(15))+(g)(sin(15))
d = (6^2) / 2(0.25)(9.8)(cos(15)) + (9.8)(sin(15))
KE = W(fr)+PE
KE = mv²/2
W(fr) =F •d=μN•d= μmg•cosα•d
PE=mgh=mg•d•sinα
mv²/2 = μmg•cosα•d+ mg•d•sinα
d=v²/2g(μ•cosα+sinα) =
= 6²/2•9.8(0.25•0.241+0.259) = 3.67 m