Hey, here's a similar question. Hopefully you can work out the differences:
Question
If the work required to speed up a car from 13 km/h to 23 km/h is 8.0×103 J, what would be the work required to increase the car’s speed from 23 km/h to 33 km/h?
Answer
First of all, I'm going to assume that the work required is 8.0 x 10^3 J. (Not really sure if it is suppose to be that or 8.0 x 103 J)
Now, to solve for this, we have to know the relationship between work and energy.
Work = Change of Kinetic Energy or Work = KE (final) - KE (initial) and KE = 1/2 mv^2
Thus, from this relationship, we can set up the equation to be:
8.0 x 10^3 J = (1/2)(m)(23km/h)^2 - (1/2)(m)(13km/h)^2
(Here, we set 23 km/h as final velocity and 13 km/h as initial velocity)
However, you have to convert the km/h to m/s to satisfy the energy unit that we are given, which is Joules. We can convert 23km/h to meters by multiplying by 1000 and convert to seconds by dividing by 3600. (1km = 1000 meters and 1 hour = 60 minutes = 60 seconds.)
And we get:
8.0 x 10^3 J = (1/2)(m)(6.4m/s)^2 - (1/2)(m)(3.6m/s)^2
Now, the only variable that we do not know is mass, which is the same for both final and inital KE since the car is the same in both cases.
8.0 x 10^3 J = 20.4m - 6.5m
After plugging into our calculator, m = 576 kg.
So, with this new mass, we can use the same work-energy (KE) equation and solve for the new work required to increase the car's speed from 23 km/h to 33 km/h.
Work = (1/2)(576)(33km/h)^2 - (1/2)(576)(23km/h)^2
Work = (1/2)(576)(9.2m/s)^2 - (1/2)(576)(6.4m/s)^2
Work = 24376.3 - 11796.5
Work = 12579.9 J
Hope that helps.
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