Answer:
255150 J
364.50233 m
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
[tex]K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}2268\times 15^2\\\Rightarrow K=255150\ J[/tex]
The initial kinetic energy of the truck is 255150 J
Acceleration is given by
[tex]a=-\dfrac{f}{m}\\\Rightarrow a=-\dfrac{700}{2268}\\\Rightarrow a=-0.30864\ m/s^2[/tex]
From equation of motion
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-15^2}{2\times -0.30864}\\\Rightarrow s=364.50233\ m[/tex]
The stopping distance of the truck is 364.50233 m
