The wind blows a 0.50 kg Styrofoam box, initially at rest, across level ground and then up an incline where it comes to a rest. The box slides 6.0 m on level ground before sliding up the 20.0

Question

Ovincline. If the effective coefficient of friction between the ground and the box is 0.60, and the wind exerts a constant, horizontally-directed force of 5.0 N, how far up the incline will the box be blown by this wind before it
comes to a stop?

Answer 1

Horizontal motion
m •a=F-μ•m•g.
a= (F-μ•m•g )/m=
=F/m - μ•g =5/0.5 – 0.6•9.8 = 4.12 m/s²

s1=at²/2 => t=sqrt(2•s/a) =1.7 s.
v=at=4.12•1.7 = 7.03 m/s.
The law of conservation of energy for incline:
mv²/2 +W(F) = W(fr) +ΔPE.
mv²/2 +F•x•cosθ = μ•m•g•cosθ•x+m•g•x•sinθ.
x= mv²/2{μ•m•g•cosθ-F•cosθ - m•g•sinθ} = 0.5 m.