In the Batmobile, Batman was chasing Joker through the streets of Gotham. As he is driving, Batman discovers that the Joker sabotaged his rid causing the engine to turn off and coast up a hill. If the Batmobile has a mass of 3900 kg and was traveling at a velocity of 40 m/s at the base of the hill, how far up the hill will the Batmobile go (assume no friction)?
Answers & Comments
Explanation:
To determine how far up the hill the Batmobile will go, we can use the principle of conservation of energy. At the base of the hill, the Batmobile has both kinetic energy and gravitational potential energy. At the top of the hill, all of the initial kinetic energy will be converted into gravitational potential energy.
The formula for gravitational potential energy is given as:
PE = m * g * h
Where:
PE is the gravitational potential energy
m is the mass of the Batmobile
g is the acceleration due to gravity (approximately 9.8 m/s²)
h is the height or distance traveled up the hill
Now, let's calculate the height (h) using the given information:
Given:
Mass of the Batmobile (m) = 3900 kg
Initial velocity (v) = 40 m/s
Acceleration due to gravity (g) = 9.8 m/s²
First, let's find the kinetic energy (KE) at the base of the hill:
KE = (1/2) * m * v²
KE = (1/2) * 3900 kg * (40 m/s)²
KE = 1/2 * 3900 * 1600 = 3,120,000 J
Now, we equate the initial kinetic energy to the gravitational potential energy at the top of the hill:
PE = KE
m * g * h = KE
h = KE / (m * g)
h = 3,120,000 J / (3900 kg * 9.8 m/s²)
h ≈ 80 meters
Therefore, the Batmobile will go approximately 80 meters up the hill before coming to a stop, assuming no frictional forces.
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