When mass M is at the position shown, it is sliding down the inclined part of a

Question

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.17 m/s. The mass stops a distance S2= 2.1 m along the level part of the slide. The distance S1= 1.23 m and the angle ? = 30.30

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.17 m/s. The mass stops a distance S2 = 2.1 m along the level part of the slide. The distance S1 = 1.23 m and the angle ? = 30.30 degree. Calculate the coefficient of kinetic friction for the mass on the surface.

Explanation / Answer

Friction acceleration on level = Cf g
Friction accel on ramp = Cf g cos 29.7 deg
Gravity on ramp = g sin 29.7
Net acceleration on ramp = g sin 29.7 - Cf g cos 29.7

Find slow down on level
d = 1/2 at^2
d = 2.15 m
a = Cf g
t = v/a
d = 1/2 v^2/Cfg
v^2 = 2 S2 Cf g
v is the velocity at the bottom of the ramp.

Now use the equation v^2 = v.o^2 + 2ax
v^2 = 2 S2 Cf g
v.o^2 = (2.39m/s)^2
x = S1
a = g sin 29.7 - Cf g cos 29.7

plug n chug

v^2 = v.o^2 + 2ax
2 S2 Cf g = (2.39m/s)^2 + 2g(sin 29.7 - Cf cos 29.7)(S1)
Cf (2 S2 g + 2g S1 cos29.7) = (2.39m/s)^2 + 2g S1 Sin 29.7
Cf = [(2.39m/s)^2 + 2g S1 Sin 29.7] / (2 S2 g + 2g S1 cos29.7)
Cf = .267

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