Answer:
a) 1350000 J
b) 1350000 J
c) 4.54m/s^2
Step-by-step explanation:
a) to find the kinetic energy of the truck you use the following formula:
![K=(1)/(2)mv^2\\\\m=3000kg\\\\v=108km/h=30m/s\\\\K=(1)/(2)(3000kg)(30m/s)^2=1350000\ J](https://img.qammunity.org/2021/formulas/physics/high-school/cere2lz1fqlgu9le7illn30o2g4184mqlk.png)
b) To stop the truck is is necessary that friction makes a work of 1350000J
c) First, you calculate the acceleration required for the truck stops 1 meter before, that is x=99m. You use the following formula:
![v^2=v_o^2-2ax\\\\v_o=30m/s\\\\x=99m\\\\v=0m/s\\\\a=(v_o^2-v^2)/(2x)=4.54(m)/(s^2)](https://img.qammunity.org/2021/formulas/physics/high-school/5z0gs35aivc3801c4kgny513cn9e550hex.png)
with this acceleration you calculate the force that stops the truck:
![F=ma\\\\F=(3000kg)(4.54(m)/(s^2))=13636.36N](https://img.qammunity.org/2021/formulas/physics/high-school/d9rr3j7phj16of2kxhieq4ikj28dgency0.png)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
TRANSLATION
a) para encontrar la energía cinética del camión utiliza la siguiente fórmula:
b) Para detener el camión es necesario que la fricción haga un trabajo de 1350000J
c) Primero, calcula la aceleración requerida para que el camión se detenga 1 metro antes, es decir x = 99m. Utiliza la siguiente fórmula:
con esta aceleración calculas la fuerza que detiene el camión: