Answer:
255150 J
364.50233 m
Step-by-step explanation:
f = Frictional force = 700 N
m = Mass of truck = 2268 kg
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
The kinetic energy is given by
![K=(1)/(2)mv^2\\\Rightarrow K=(1)/(2)2268* 15^2\\\Rightarrow K=255150\ J](https://img.qammunity.org/2020/formulas/physics/high-school/duuwshdr45lhh5gqfmvnzaoemruhr6wb9n.png)
The initial kinetic energy of the truck is 255150 J
Acceleration is given by
![a=-(f)/(m)\\\Rightarrow a=-(700)/(2268)\\\Rightarrow a=-0.30864\ m/s^2](https://img.qammunity.org/2020/formulas/physics/high-school/y39j7i9dxh7bjcwsf6gigu2e0ie7pc2y44.png)
From equation of motion
![v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-15^2)/(2* -0.30864)\\\Rightarrow s=364.50233\ m](https://img.qammunity.org/2020/formulas/physics/high-school/t028dvqok320ff52fq45erkuxda60u2uh4.png)
The stopping distance of the truck is 364.50233 m