Question

2A constant-magnitude horizontal force is applied to a body of mass m= 5.0 kg sliding along horizontal surface. The force will cause the speed of body change from v0 =3.0m/s to v1 = 7.5m/s during a time interval of t1 = 2.0s. Thereafter the interaction associated with the applied force disappears and the body will stop 1.0 seconds later. Determine the magnitude of constant force applied during the time interval t 1.
please help.

Answer 1

a=(v2-v1)/t.
Acceleration is defined as the rate of change of velocity with respect to time.
a=(7.5-3)/2 m/s^2
a=2.25m/s^2
now by formula
F=ma
F=5*2.25 N
F=11.25 N

Answer 2

I answered your questions in the first posting of this question.
The friction only acts for 1 second, so the deceleration is a = -7.5 m/s^2
Then, 'adding' up the forces, I already put the minus sign for the friction. So, maybe you used a minus sign twice.
External force - frictional force = ma Keep it negative in this equation, so you add it to ma.