Answer:
Explanation:
a is the acceleration
μ is the coefficient of friction
Acceleration of the object is given by
[tex]a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2[/tex]
Velocity at the bottom
[tex]v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s[/tex]
after travelling 4m , its velocity becomes 0
[tex]a=\frac{v^2-u^2}{2s}[/tex]
[tex]a=\frac{0-u^2}{2s}[/tex]
[tex]a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2[/tex]
Coefficient of kinetic friction
μ = F/N
[tex]=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31[/tex]
Therefore, the Coefficient of kinetic friction is 0.31
Answer:
Coefficient of kinetic friction = 0.31
Explanation:
g = 10 m/s²
distance covered up the hill, S = 8 m
Horizontal distance covered, d = 4 m
From the free body diagram attached:
[tex]ma = mgsin \theta - \mu_{k} N\\N = mgcos \theta\\ma = mgsin \theta - \mu_{k} mgcos \theta\\a = gsin \theta - \mu_{k} gcos \theta\\a = 10sin 30 - (0.4*10cos30)\\a = 1.4 m/s^2[/tex]
The velocity at the bottom can be calculated by:
[tex]v^{2} = u^{2} + 2aS\\u = 0 m/s\\v^{2} = 2aS\\v^{2} = 2*1.54*8\\v^{2} = 24.64\\v = 4.96 m/s[/tex]
Work done through to frictional force = change in kinetic energy:
[tex]f_{k} d = 0.5 mv^{2} \\\mu_{k} mgd = 0.5mv^{2} \\\mu_{k} = \frac{0.5v^{2} }{gd} \\\mu_{k} = \frac{0.5*4.96^{2} }{10*4} \\\mu_{k} = 0.31[/tex]
