The 70 kg crate shown in figure falls horizontally from a truck which is travelling at 70 km/h. If the crate slides 50 m on the ground without tumbling before coming to rest, determine the coefficient of friction between the crate and the road. Take initial speed of the crate as the speed of the truck.

a. 0.345
b. 0.385
c. 0.355

d. 0.375

Consider the horizontal forces acting on the crate.
The net horizontal force acting on the crate is F which is equal to m a.
but f = m N and N = m g
where m is the mass of the crate.

m m g = m a
=> a = m g

where a is the acceleration of the crate

The initial velocity of the crate can be written as
u2 = - 2 a s
where u is the initial velocity of the crate and s is the distance travelled by the crate.
a = - u2/(2 s) = - 3.78 m/s2
The negative sign indicates retardation
=> m = a/g = 0.385.

The coefficient of friction between the crate and the road is 0.385.