When a car's speed doubles from 20 m/s to 40 m/s, by how many times does the car's kinetic energy increase? Show your work. The mass of the car is 1000 kg.
k=1/2mv^2
2^2 = 4
To find how many times the car's kinetic energy increases, we need to compare the initial kinetic energy (K1) to the final kinetic energy (K2) using the formula K = 1/2 * m * v^2.
Given that the initial speed (v1) is 20 m/s, the final speed (v2) is 40 m/s, and the mass (m) is 1000 kg, our calculations are as follows:
For the initial speed (v1 = 20 m/s):
K1 = 1/2 * m * v1^2
K1 = 1/2 * 1000 kg * (20 m/s)^2
K1 = 1/2 * 1000 kg * 400 m^2/s^2
K1 = 200,000 kg * m^2/s^2
For the final speed (v2 = 40 m/s):
K2 = 1/2 * m * v2^2
K2 = 1/2 * 1000 kg * (40 m/s)^2
K2 = 1/2 * 1000 kg * 1600 m^2/s^2
K2 = 800,000 kg * m^2/s^2
Now, let's calculate the ratio of K2 to K1 to determine how many times the car's kinetic energy increases:
Ratio = K2 / K1
Ratio = (800,000 kg * m^2/s^2) / (200,000 kg * m^2/s^2)
Ratio = 4
Therefore, the car's kinetic energy increases by a factor of 4, or in other words, the car's kinetic energy doubles.
To calculate the increase in kinetic energy, we need to compare the initial kinetic energy to the final kinetic energy.
First, let's calculate the initial kinetic energy (K1) when the car's speed is 20 m/s.
K1 = (1/2) * m * v^2
= (1/2) * 1000 kg * (20 m/s)^2
= (1/2) * 1000 kg * 400 m^2/s^2
= 200,000 joules
Now, let's calculate the final kinetic energy (K2) when the car's speed doubles to 40 m/s.
K2 = (1/2) * m * v^2
= (1/2) * 1000 kg * (40 m/s)^2
= (1/2) * 1000 kg * 1600 m^2/s^2
= 800,000 joules
To find the increase in kinetic energy, we subtract the initial kinetic energy from the final kinetic energy:
Increase in kinetic energy = K2 - K1
= 800,000 joules - 200,000 joules
= 600,000 joules
So, the car's kinetic energy increases by 600,000 joules when its speed doubles from 20 m/s to 40 m/s.