Question

When a steel block is released from rest on a steep inclined plane, it moves 50.0 cm in 0.404 s. The inclined plane makes an angle of 50.00 with respect to the horizontal. There is a constant force of kinetic friction between the steel block and the inclined surface. What is the kinetic coefficient of the frictional force?

Answer 1

$N=mg\cos\alpha$

$mg\sin\alpha-\mu mg\cos\alpha=ma$

$S=\frac{at^2}{2}\to a=\frac{2S}{t^2}=\frac{2\cdot0.5}{0.404^2}=6.13m/s^2$

$\mu=\tan\alpha-\frac{a}{g\cos\alpha}=\tan50°-\frac{6.13}{9.81\cdot\cos50°}=0.23$ Answer.