Answer:
Approximately [tex]63\; \rm m[/tex].
Explanation:
Convert the initial speed of this truck to standard units:
[tex]\begin{aligned} v &= 72\; \rm km \cdot h^{-1} \times \frac{1\; \rm h}{3600\; \rm s} \times \frac{1000\; \rm m}{1\; \rm km} \\ &= 20\; \rm m \cdot s^{-1}\end{aligned}[/tex].
Calculate the initial kinetic energy of this truck:
[tex]\begin{aligned}\text{KE} &= \frac{1}{2} \, m \cdot v^{2} \\ &= \frac{1}{2} \times 2600\; \rm kg \times (20\; \rm m \cdot s^{-1})^{2} \\ &= 5.2 \times 10^{5} \; \rm J\end{aligned}[/tex].
The kinetic energy of the truck would be [tex]0[/tex] when the truck is not moving. Hence, the friction on the truck would need to do [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work on the truck to bring the truck to a stop.
Calculate the displacement required to achieve [tex](-5.2 \times 10^{5}\; \rm J)[/tex] of work at [tex]8200\; \rm N[/tex]:
[tex]\begin{aligned}x &= \frac{W}{F} \\ &= \frac{-5.2 \times 10^{5}\; \rm J}{8200\; \rm N} \approx -63\; \rm m\end{aligned}[/tex].
This result is negative because the direction of friction is opposite to the direction of the motion of the truck.
Hence, the truck would come to a stop after skidding for approximately [tex]63\; \rm m[/tex].
