m = 2.5 kg, α = 30º, s =2, k = 0.2
ΔPE =KE +W(fr)
ΔPE = m•g•Δh =m•g•s•sinα
KE = m•v²/2,
W(fr) = k•N•s= k• m•g•cosα•s
m•g•s•sinα = m•v²/2 + m•g•cosα•s,
v= sqrt{2•g•s(sinα –k• cosα} =
= sqrt{2•9.8•2(0.5 -0.2•0.866)} =
=3.58 m/s
A 2.5 kg wooden block slides from rest down an inclined plane that makes an angle of 30o with the horizontal. If the plane has a coefficient of kinetic friction of 0.20, what is the speed of the block after slipping a distance of 2.0 m?
1 answer