A car is moving such that its kinetic energy is 6.20×105 J. The car accelerates and the engine does 2.90×105 J of work to accelerate the car. What is the final kinetic energy of the car? Ignore energy lost to friction.
(Your answer must be accurate to at least 1%. Give your answer to at least three significant figures.)
Just add the two.
To solve this problem, we need to use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
In this case, the work done by the engine is 2.90×10^5 J, and we need to find the final kinetic energy. Let's denote the initial kinetic energy as K_i and the final kinetic energy as K_f.
According to the work-energy theorem, the change in kinetic energy is given by the formula:
ΔK = K_f - K_i = work done
Since we know the work done (2.90×10^5 J) and the initial kinetic energy (6.20×10^5 J), we can substitute these values into the equation:
K_f - 6.20×10^5 J = 2.90×10^5 J
To isolate K_f, we can add 6.20×10^5 J to both sides of the equation:
K_f = 2.90×10^5 J + 6.20×10^5 J
K_f = 9.10×10^5 J
Therefore, the final kinetic energy of the car is 9.10×10^5 J.