# Calculating final kinetic energy from change in velocity by percent

#### pmt

The kinetic energy of a car is 1.02 × 105 J. If the car s speed increases by 28.2%, then what is the new kinetic energy of the car?

I set up the ratio Kf/Ki=Vf2/Vi2
Vf= 1 + 28.2/100)2Vi2
this cancels the right side of the equation out to (1.282)2
to find Kf, I then multiplied both sides by Ki, making the left side (1.282)2X10,200
this means Kf=16764J

Why is this incorrect?

Last edited:

#### skeeter

The kinetic energy of a car is 1.02 × 105 J. If the car s speed increases by 28.2%, then what is the new kinetic energy of the car?
I assume $KE_0 = 1.02 \times 10^5 \, J$ ...

$KE_0 = \dfrac{1}{2}mv_0^2$

$KE_f = \dfrac{1}{2}m(1.282v_0)^2$

$\dfrac{KE_f}{KE_0} = \dfrac{\frac{1}{2}m(1.282v_0)^2}{\frac{1}{2}mv_0^2}$

$\dfrac{KE_f}{KE_0} = \dfrac{(1.282v_0)^2}{v_0^2} = 1.282^2 \implies KE_f = KE_0(1.282^2) = 1.68 \times 10^5 \, J$

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