We know that the work is related to the kinetic energy by the work energy theorem:
![W=K_f-K_i](https://img.qammunity.org/2023/formulas/physics/college/1dqyj15gkqdq44obqgf2d083n918iumge4.png)
In this case the final velocity of the car is 0 m/s; with this in mind and plugging the values given for the initial velocty we have:
![\begin{gathered} W=(1)/(2)(1180)(0)^2-(1)/(2)(1180)(18.2)^2 \\ W=-195431.6 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/4lk87kk2fulvs68fz1sij5wuio8bmvdktb.png)
Now, we know that the work is given by:
![W=Fd](https://img.qammunity.org/2023/formulas/physics/high-school/hpc1le8wrnyqcrfx7se11dqmtkccnojf2k.png)
plugging the distance and the work done we have:
![\begin{gathered} -195431.6=0.411F \\ F=-(195231.6)/(0.411) \\ F=-4.76*10^5 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/34ki0x7rfoniymzev5870twd17n4rir8zg.png)
Since we need the magnitude we drop the sign, therefore the magnitude of the force is:
![F=4.76*10^5\text{ N}](https://img.qammunity.org/2023/formulas/physics/college/zvxzj5tje0hjqldb3skm0k02mzurn4ish3.png)