HELP!!!! FAST!!!! 20 POINTS!!!! You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equalmass (3,000 kg each) were involved in a rear-end accident at a stop sign. Here is what you know:
• Truck 1 started from rest at the top of a 22-meter hill and coasted toward the intersection.
.
Truck 2 was on a flat stretch of road at the bottom of the hill directly in front of Truck 1.
At the bottom of the hill, Truck 1 was going 20 m/s; Truck 2 was going 35 m/s.
.
There were no skid marks left by Truck 2. The collision occurred at the stop sign where Truck 2 had
stopped
• After the collision, both trucks moved together in the same direction at 10 m/s before slowly rolling
to a stop.
You must now push Truck 2, using 1,000 N of force, 8 meters off to the side of the road so no one
else gets hurt.
1. What was the potential energy of Truck 1 at the top of the hill? 2. What was the kinetic energy of each truck before the began braking? 3. A. How much energy did truck 1 lose from the top to the bottom of the hill? B. Where do you suppose that energy went?

Answer:

1) Kinetic energy of the truck 1 is 600,000 joules.

Kinetic energy of the truck 2 is 1,837,500 joules.

2 a)Potential energy of truck 1 is 646,800 Joules.

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

Step-by-step explanation:

Mass of the truck 1 and truck 2,m = 3000 kg

Velocity of the truck 1,  = = 20 m./s

Velocity of the truck 2, = 35 m./s

1. Kinetic energy of the truck 1:

Kinetic energy of the truck 2:

2.a) Energy loose by the truck 1:

Initial height of the truck on the hill = 22 m

At that height potential energy will be maximum:

Potential energy of truck 1 at that height =  

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

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Final answer:

The potential energy of Truck 1 at the top of the hill is 646,800 J. The kinetic energy of each truck before they began braking is 600,000 J for Truck 1 and 3,712,500 J for Truck 2. The energy lost by Truck 1 from the top to the bottom of the hill is 646,800 J, which is converted into kinetic energy and transferred to Truck 2 during the collision.

Explanation:

1. The potential energy of Truck 1 at the top of the hill can be calculated using the formula PE = mgh, where m is the mass of the truck, g is the acceleration due to gravity, and h is the height of the hill. Plugging in the values, PE = (3,000 kg)(9.8 m/s^2)(22 m) = 646,800 J.

2. The kinetic energy of each truck before they began braking can be calculated using the formula KE = 0.5mv^2, where m is the mass of the truck and v is the velocity. For Truck 1, KE = (0.5)(3,000 kg)(20 m/s)^2 = 600,000 J. For Truck 2, KE = (0.5)(3,000 kg)(35 m/s)^2 = 3,712,500 J.

3.A. The energy that Truck 1 lost from the top to the bottom of the hill can be calculated by subtracting the potential energy at the top from the kinetic energy at the bottom. Therefore, energy lost = PE(top) - KE(bottom) = 646,800 J - 0 J = 646,800 J.

3.B. The energy is converted into kinetic energy and transferred to Truck 2 when they collide. The collision between the two trucks results in an increase in kinetic energy for Truck 2 and a decrease in kinetic energy for Truck 1.

Learn more about Energy Transfer here:

brainly.com/question/32491696

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