For the 70 kg crate shown in the figure, determine its velocity in 5 s starting from rest when P = 500 N and coefficient of friction between the crate and the horizontal plane is 0.2.

a. 18.5 m/s
b. 19.5 m/s
c. 20.5 m/s
d. 21.5 m/s

Consider the horizontal forces acting on the crate
The net horizontal force acting on the crate is equal to m a

500 cos 450 - 0.2 N = 70 a ------- (1) 
as f = mk NR

Consider the vertical forces acting on the crate
The net force acting on the crate in the vertical direction is given as
NR - 686.7 + 500 sin
 450 = 0 ------- (2)
By solving (2), we get NR = 333.15 N
Substituting this value in (1) gives a = 4.1 m/s2

The acceleration of the crate is constant since the applied force is constant.
As the crate starts from rest, the velocity of the crate in 5 s is given by
v = a t = 4.1 (5) = 20.5 m/s

The velocity of the crate in 5 s starting from rest is
20.5 m/s.