Question

Jim is driving a 2268-kg pickup truck at 15.0 m/s and releases his foot from the accelerator pedal. The car eventually stops due to an effective friction force that the road, air, and other things exert on the car. The friction force has an average magnitude of 700 N . Part A Determine the initial kinetic energy of the truck.
Part B Determine the stopping distance of the truck.

Answer 1

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

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}2268\times 15^2\\\Rightarrow K=255150\ J$

The initial kinetic energy of the truck is 255150 J

Acceleration is given by

$a=-\dfrac{f}{m}\\\Rightarrow a=-\dfrac{700}{2268}\\\Rightarrow a=-0.30864\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-15^2}{2\times -0.30864}\\\Rightarrow s=364.50233\ m$

The stopping distance of the truck is 364.50233 m