A 1250 kg car slams on the brakes and skids on the icy road. If the car has to apply just 562N to get moving again, then 250N to maintain a constant speed (assume air resistance is negligible), what are the coefficients of static friction and kinetic friction of the road.
4 years ago
Answered By Majid B
a) Start to move:
$F_{app}=F_{fs}=\mu_s.F_N=\mu_s.F_g=\mu_s.m.g$Fapp=Fƒ s=μs.FN=μs.Fg=μs.m.g => $\mu_s=\frac{\left(F_{app}\right)}{m.g}=\frac{\left(562N\right)}{\left(1250kg\right)\left(9.81\left(\frac{m}{s^2}\right)\right)}=0.046$μs=(Fapp)m.g =(562N)(1250kg)(9.81(ms2 )) =0.046
b) Maintain a constant speed:
$F_{app}=F_{fk}=\mu_k.F_N=\mu_k.F_g=\mu_k.m.g$Fapp=Fƒ k=μk.FN=μk.Fg=μk.m.g => $\mu_k=\frac{\left(F_{app}\right)}{m.g}=\frac{\left(250N\right)}{\left(1250kg\right)\left(9.81\left(\frac{m}{s^2}\right)\right)}=0.020$μk=(Fapp)m.g =(250N)(1250kg)(9.81(ms2 )) =0.020