a)
From the information given,
the puck started from rest. This means that
initial velocity = 0
Force = 50N
Mass = 2 kg
Recall,
Force = mass x acceleration
acceleration = Force/mass = 50/2 = 25m/s
The acceleration with which the ice moved after the force was removed = 25 m/s
The velocity of the puck after the force was removed is the final velocity, v
If distance travelled from rest is 0.32, we would find the final velocity by applying the formula
v^2 = u^2 + 2as
where
s = distance = 0.32
u = initial velocity = 0
Thus,
v^2 = 0^2 + 2 x 25 x 0.32 = 16
v = square root of 16
v = 4 m/s
The velocity of the puck after the force was removed = 4m/s
b) The puck travelled at a constant speed of 4m/s after the force was removed. This means that the acceleration did not change.
distance travelled = 15m
After travelling 15m, it moved into a rough part with a frictional force of 12N.
When travelling through the rough path,
net force = 50 - 12 = 38N
acceleration in the rough path is
acceleration = 38/2 = 19m/s^2
Since the puck came to a stop, final velocity, v = 0
initial velocity in the rough path = 4m/s
We would calculate the distance travelled in the rough path by applying the formula,
v^2 = u^2 + 2as
By substituting the values,
0^2 = 4^2 + 2 x - 19 x s
The negative sign shows that the puck was decelerating
0 = 16 - 38s
38s = 16
s = 16/38
s = 0.42m
Distance travelled in the rough path = 0.42m
Total distance = 0.32 + 15 + 0.42 = 15.74 m