Answer
$F_{avg}=6.5\times 10^3 N$
Work Step by Step
We can find the required average force as follows:
Initial momentum of the tractor-trailer is given as
$P_1=m v_1$
$\implies P_1=5000 Kg\times 0.833m/s$ ($As v_1=3Km/h=0.833m/s$)
This simplifies to:
$P_1=4165m/s$
As the final velocity is zero , therefore $P_2=0$
Now $F_{avg}=\frac{P_2-P_1}{\Delta t}$
We plug in the known values to obtain:
$F_{avg}=\frac{0-4165}{0.64}$
This simplifies to:
$F_{avg}=6.5\times 10^3 N$
