Question

Jim is driving a 2268-kg pickup truck at 22 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 900 N. Part A Determine the stopping distance of the truck. Express your answer with the appropriate units.

Answer 1

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

$E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J$

The work done by the friction force is given by:

$W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m$