A 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.

Question

A 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.

anonymous · Answer

First figure out how much frictional force is acting on it. The deceleration from friction can be found from: V2 = V1 + at
for the last second with V2 = 0.
Then you can find the frictional force from F=ma
Then find the acceleration from the external force, same way as for frictional force. Use that acceleration to find the external force:
External force - frictional force = ma

anonymous · Answer

I have found friction force to be 18.75 KN, if i use this value to find acceleration from the external force i get the same value, i am some how confused with this. The first deceleretion i got is -3.5 m/s/s , somebody help

anonymous · Answer

Anybody help, i couldnt get through this yet

anonymous · Answer

If you still need help with this:
Find the external force acceleration from V2 = V1 + at
using the given initial and final velocities and the time the force is applied, t = 2 sec.
This is actually the combined external and frictional force acceleration because friction is also acting on it.
So, use this acceleration to find the external force by subtracting the frictional force:
External force - frictional force = ma

anonymous · Answer

i get -7.5kN i dont know why but i think it wrong. the combined ext force and friction force acceleration is 2.25m/s/s i believe that should be right as i followed procedure. then i went on to find deceleration at 1s when v2=0 and t=2s i got -3.75, i used this value to get friction force from f=ma, i got -18.75kN. then exforce-friction force=ma as u said. where do i go wrong mate. i need this help big time. i dont like to fail exams.

anonymous · Answer

The value for acceleration for the combined forces is correct: 2.25 m/s^2
The deceleration occurs in 1 second so a = -7.5 m/s^2
This give a frictional force of -37.5 N
For the final calculating of the external force, you may have used a negative sign twice.
External force - 37.5 N = ma
External force = 37.5N + (5 kg)(2.25 m/s^2) = 48.75 N