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The total mechanical energy of the car/rider combination is 65162.5J

at the top of the hill. What is the kinetic energy of the car/rider
combination at the bottom of the hill? (Ignore friction and air resistance)
Round to the nearest tenth.

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User Jassiel
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1 Answer

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The law of conservation of energy states that the total energy of a system remains constant if no external work is done on the system. Therefore, the total mechanical energy of the car/rider combination at the top of the hill must be equal to the total mechanical energy of the car/rider combination at the bottom of the hill.

At the top of the hill, the total mechanical energy of the car/rider combination is:

E_top = mgh

where m is the mass of the car/rider combination, g is the acceleration due to gravity, and h is the height of the hill.

At the bottom of the hill, the total mechanical energy of the car/rider combination is:

E_bottom = 1/2mv^2

where m is the mass of the car/rider combination and v is the velocity of the car/rider combination at the bottom of the hill.

We can set these two expressions equal to each other:

mgh = 1/2mv^2

Simplifying and solving for v, we get:

v = sqrt(2gh)

where sqrt represents the square root function.

Substituting the given values, we have:

v = sqrt(2*9.8*65162.5)

v = 359.2 m/s

Therefore, the kinetic energy of the car/rider combination at the bottom of the hill is:

E_kinetic = 1/2mv^2

E_kinetic = 1/2*80*(359.2)^2

E_kinetic = 5,200,179.2 J

Rounding to the nearest tenth, the kinetic energy of the car/rider combination at the bottom of the hill is approximately 5,200,179.2 J.
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User Mattgately
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