Answer
given,
mass of block (m)= 6.4 Kg
spring is stretched to distance, x = 0.28 m
initial velocity = 5.1 m/s
a) computing weight of spring
k x = m g


k = 224 N/m
b) 




c) 


d) 


e)


A = 0.682 m
Force =
=
F = 94.20 N
Answer:
God is omnipresent.
Explanation:
This means God is everywhere and He works where ever we are in the world
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:

The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg

Now the total force required is:
0.0702N+0.803N=0.873N