Question

Jim is driving a 2268-kg pickup truck at 30.0m/sand 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 700N .Part A Determine the initial kinetic energy of the truck.Part B Determine the stopping distance of the truck.

Answer 1

Explanation:

Given that,

Mass of the truck, m = 2268 kg

Speed of the truck, v = 30 m/s

The friction force has an average magnitude of 700 N, f = -700 N

(a) The initial kinetic energy of the truck is given by :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2268\times (30)^2\\\\K=1.02\times 10^6\ J$

(b) Finally, the stops due to an effective friction force that the road, final speed is 0. Let d is the stopping distance of the truck. using third equation of motion to find it as :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}$

Since, f = ma

$d=\dfrac{-u^2m}{2f}\\\\d=\dfrac{-(30)^2\times 2268}{-2\times 700}\\\\d=1458\ m$

So, the stopping distance of the truck is 1458 meters.