To solve the problem, we can go through the following steps:

a) To calculate the acceleration (a) of the object, we can use the formula of motion. Since the object moves 4m in 2 seconds, we can find the acceleration using the formula:

s=ut+12at2s=ut+21​at2
Where:

s=4 ms=4m (displacement)
u=0 m/su=0m/s (initial velocity, it starts from rest)
t=2 st=2s
Substituting these values in:

4=0×2+12a(22)4=2aa=2 m/s24=0×2+21​a(22)4=2aa=2m/s2
b) To find the frictional force (F_friction), we first calculate the net force (F_net) acting on the object:

Fnet=m⋅aFnet​=m⋅a
The mass (m) is 800g, which is 0.8kg. Therefore:

Fnet=0.8 kg⋅2 m/s2=1.6 NFnet​=0.8kg⋅2m/s2=1.6N
The total force pulling the object is 2N (Fk), so to find the force of friction:

Fk=Fnet+FfrictionFfriction=Fk−Fnet=2 N−1.6 N=0.4 NFk​=Fnet​+Ffriction​Ffriction​=Fk​−Fnet​=2N−1.6N=0.4N
c) After 5 seconds, the force pulling is stopped. We can calculate the remaining distance the object travels before stopping. Using the formula:

v=u+atv=u+at
Where the total time (t) from the start is 5s:

v=0+2⋅5=10 m/s(this is the speed at the end of 5 seconds)v=0+2⋅5=10m/s(this is the speed at the end of 5 seconds)
When the force is no longer applied, the object will decelerate due to friction:

a=−Ffrictionm=−0.40.8=−0.5 m/s2a=−mFfriction​​=−0.80.4​=−0.5m/s2
Now, we will find the distance (d) it takes to stop using the equation:

v2=u2+2adv2=u2+2ad
Setting v=0v=0:

0=(10)2+2⋅(−0.5)⋅d0=100−dd=100⋅0.5=50 m0=(10)2+2⋅(−0.5)⋅d0=100−dd=100⋅0.5=50m
Thus, the object will travel an additional 50m after the force is removed before stopping.