A) What is the kinetic energy of a 1200kg car traveling at a speed of 30 m/s (≈65mph)?
B) From what height would the car have to be dropped to have this same amount of kinetic energy just before impact?
Use
Ek=(1/2)mv²
Equate
Ek=mgh
h=height
540,000
To calculate the kinetic energy of the car, we can use the formula:
Kinetic Energy (KE) = 0.5 * mass * velocity^2
A) First, we need to calculate the kinetic energy using the given mass (m = 1200 kg) and velocity (v = 30 m/s):
KE = 0.5 * 1200 kg * (30 m/s)^2
KE = 0.5 * 1200 kg * 900 m^2/s^2
KE = 540,000 Joules
Therefore, the kinetic energy of the car is 540,000 Joules.
B) To find the height from which the car would have to be dropped to have the same kinetic energy just before impact, we can use the principle of conservation of energy. The potential energy gained from the height should be equal to the kinetic energy just before impact.
Potential Energy (PE) = Mass * gravitational acceleration * height
Since the potential energy gained from the height will be converted into kinetic energy just before impact, we can equate the two:
PE = KE
Mass * gravitational acceleration * height = 540,000 Joules
Here, the mass can be written as m = 1200 kg, and the gravitational acceleration can be considered as g = 9.8 m/s^2.
Using these values, we can rearrange the equation to solve for the height:
1200 kg * 9.8 m/s^2 * height = 540,000 Joules
height = 540,000 Joules / (1200 kg * 9.8 m/s^2)
height ≈ 45.92 meters
Therefore, the car would have to be dropped from a height of approximately 45.92 meters to have the same kinetic energy just before impact.