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Grade 12th passMechanics

An 1100kg car is travelling at 46km per hour on level road.the brakes are applied long enough to remove 51kj of kinetic energy find final speed

Profile image of Hamdia
7 Years agoGrade 12th pass
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1 Answer

Profile image of Khimraj
7 Years ago
I started out by using the first equation, abbreviated as:
W = KE(f) - KE(i)
I then converted 51 kJ into 51,000 J and put that value in for "work."
After that, I substituted the KE values for (1/2)mv^2 according to the second equation I listed.
51,000 J = [(1/2)mv^2] - [(1/2)mv^2]
I plugged in some more values. In order to get the units to come out as Joules, I had to convert 46 km/h for the initial veloctity to 12.7777... m/s.
51,000 J = [(1/2)*(1,100 kg)*(v^2)] - [(1/2)*(1,100 kg)*(12.777777... m/s)^2]
51,000 J = [550 kg*v^2] - [7027.77777... J]
58027.777... J = [550 kg*v^2]
v^2=105.5050505...
v=10.27...
v=10 m/s