First, we need to find the speed of the crate at the end of the rough patch. We can use conservation of energy:
initial potential energy = final kinetic energy + final potential energy + work done by friction
mgh1 = (1/2)mv^2 + mgh2 + Ffriction * d
where m = mass of crate, h1 = initial height, v = velocity at end of rough patch, h2 = final height, Ffriction = force of friction, and d = distance of rough patch.
The force of friction is given by Ffriction = μk * m * g, where g is the acceleration due to gravity. Plugging in the given values and solving for v, we get:
v = 4.36 m/s
Next, we can analyze the motion of the crate up the incline using conservation of energy again:
initial kinetic energy = final potential energy + work done by friction
(1/2)mv^2 = mgh2 + Ffriction * d'
where d' is the distance along the incline. We can find this distance using trigonometry: d' = d/sinθ, where θ = 25°. Plugging in the given values and solving for h2, we get:
h2 = 1.54 m
A 6.0 kg crate slides from rest at h1 = 1.46 m down to a horizontal surface where it passes over a 1.60 m rough patch. The rough patch has a kinetic friction coefficient (μk) of 0.34. An incline of 25.0° follows, as indicated in the diagram below. What maximum height, h2, does the crate reach?
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