Answer:
[tex]\mu = 0.6125[/tex]
Explanation:
given,
height of the inclined plane = h
length of the inclined plane = 8 m
angle made with the horizontal= 30°
stopped in the rough horizontal surface = 2 m
kinetic friction coefficient = 0.4
acceleration due to gravity = 10 m/s²
coefficient of kinetic friction on horizontal surface = ?
using energy conservation
[tex]mgh = \dfrac{1}{2}mv^2 + f_k N l[/tex]
[tex]\dfrac{1}{2}mv^2= mgh - \mu_k (mg cos \theta)l[/tex]
[tex]\dfrac{1}{2}v^2= gl sin\theta - \mu_k (g cos \theta)l[/tex]
[tex]\dfrac{1}{2}v^2= 9.8\times 8\times 0.5 - 0.4\times (9.8 cos 30^0)\times 8[/tex]
[tex]v = \sqrt{2 \times 12.041 }[/tex]
v = 4.91 m/s
[tex]\dfrac{1}{2}mv_i^2= \dfrac{1}{2}mv^2 + f_k N l[/tex]
[tex]\dfrac{1}{2}v_i^2= 0 + (\mu g) l[/tex]
[tex]4.9^2\times 0.5 = \mu \times 9.8 \times 2[/tex]
[tex]\mu = 0.6125[/tex]
