a) 
We can find the acceleration of the box by using the suvat equation:

where
s is the distance travelled
u is the initial velocity
t is the time
a is the acceleration
For the box in the problem,
s = 16 m
t = 9 s
u = 0 (it starts from rest)
Solving for a, we find the acceleration:

C) Tension force in the chain: 446 N
In order to find the normal force, we have to write the equation of the forces along the vertical direction.
We have 3 forces acting along this direction on the box:
- The normal force, N, upward
- The force of gravity, mg, downward (where m= mass of the box and g = acceleration of gravity)
- The vertical component of the tension of in the chain,
, upward
So the equation of the forces along the vertical direction is
(1)
Along the horizontal direction, instead, we have the following equation:
(2)
where
is the horizontal component of the tension in the chain
is the frictional force
a is the acceleration
From (1) we write

And substituting into (2),

And substituting:
m = 100 kg




We find the tension in the chain:

B) Normal force: 792 N
We can now find the normal force by using again equation (1):
And substituting:
T = 446 N
m = 100 kg


We find:
