is the work needed to bring a car's speed from 0 to 10 km/h less than, equal to or more then the work needed to bring its speed from 10 to 20 km/h? if the amounts of work are different, what is the ratio between them?
ke=1/2 (mass)(velocity)
ke=1/2mv^2
ke= 1/2m(10^2-0^2) ke=1/2m(20^2-10^2)
ke=1/2m(100) ke=1/2m(400-100)
ratio is 300/100 or 3/1
To determine the work needed to bring a car's speed from 0 to 10 km/h and from 10 to 20 km/h, you need to consider the concept of kinetic energy.
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Since kinetic energy is directly proportional to the square of the velocity, we can use this relationship to calculate the work done.
Let's denote the work done in the two scenarios as Work1 and Work2, respectively.
1. Work1: Bringing the car from 0 to 10 km/h
The initial velocity (v1) is 0 km/h, and the final velocity (v2) is 10 km/h. To find the work done, we need to calculate the change in kinetic energy between these two speeds.
Kinetic energy at v1: KE1 = (1/2) * m * (v1^2)
Kinetic energy at v2: KE2 = (1/2) * m * (v2^2)
The change in kinetic energy is ΔKE1-2 = KE2 - KE1
= (1/2) * m * (v2^2) - (1/2) * m * (v1^2)
= (1/2) * m * [ (v2^2) - (v1^2) ]
2. Work2: Bringing the car from 10 to 20 km/h
Similarly, the initial velocity (v1') is 10 km/h, and the final velocity (v2') is 20 km/h.
Kinetic energy at v1': KE1' = (1/2) * m * (v1'^2)
Kinetic energy at v2': KE2' = (1/2) * m * (v2'^2)
The change in kinetic energy is ΔKE1'-2' = KE2' - KE1'
= (1/2) * m * (v2'^2) - (1/2) * m * (v1'^2)
= (1/2) * m * [ (v2'^2) - (v1'^2) ]
Now, let's compare the two works:
Comparing Work1 and Work2: (ΔKE1-2) / (ΔKE1'-2') = [(1/2) * m * (v2^2 - v1^2)] / [(1/2) * m * (v2'^2 - v1'^2)]
= (v2^2 - v1^2) / (v2'^2 - v1'^2)
Substituting the given values v1 = 0 km/h, v2 = 10 km/h, v1' = 10 km/h, and v2' = 20 km/h, we get:
[(10^2) - (0^2)] / [(20^2) - (10^2)] = 100 / (400 - 100) = 100 / 300 = 1/3
Therefore, the ratio between the work needed to bring the car's speed from 0 to 10 km/h and from 10 to 20 km/h is 1:3 or 1/3. This implies that the work needed to bring the car's speed from 10 to 20 km/h is three times greater than the work needed to bring it from 0 to 10 km/h.