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3 years ago

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Jim is driving a 2268-kg pickup truck at 30.0m/sand releases his foot from the accelerator pedal. The car eventually stops due t

o 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.

1 answer:

Ronch [10]3 years ago

3 0

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.

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