a) a truck tilts its flat bed slowly to dispose of a 95.0 kg crate. For small angles the crate stays put but just begins to slide when the angle of tilt is 24.0°. Make a free body diagram for the crate when it is on the verge of sliding, clearly showing the relevant forces and calculate the coefficient of static friction between the crate and the truck’s bed.
b)If the bed is tilted at 30.0°, what will be the acceleration of the crate if coefficient of kinetic friction is 0.35?
c)If the crate slides 1.5 m from rest, what is its velocity at the bottom for case b)?
The Attempt at a Solution
a)The coefficient friction at the point where the crate just starts sliding = tan (the angle) = tan 24 = .445
b)The force of gravity perpendicular to the plane is called the normal force and it is equal to mg cos(angle) = (95)(9,8)cos(3) = 806.246
the force of gravity parallel to the plane is
mg sin(angle) = (95)(9.8)sin(3) = 465.5
force of friction parrellel to plane(Ff) = uFn = (.3)(806.246) = 241.872
the net force = Fn – Ff = 4.65 – 241.872 = 223.629
divide this by the original mass for acceleration = 223.629/95 = 2.354
c) i dont know how to approach this problem...