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?
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