Question

A car traveling at 60 mph has how much more energy than a car going at 20 mph? How many times does the kinetic energy of a car increase when traveling 60 mph as opposed to traveling 30 mph? K.E. increases _____ times.

Answer 1

K.E. increases by 9 times

Explanation:

The kinetic energy of a car is given by:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the car

v is its speed

From this definition, we see that the kinetic energy depends on the square of the velocity. Assuming that both cars have same mass, m, the kinetic energy of the first car is:

$K_1 = \frac{1}{2}m(60)^2$

while the kinetic energy of the second car is

$K_2 = \frac{1}{2}m(20)^2$

if we calculate the ratio, we get

$\frac{K_1}{K_2}=\frac{(60)^2}{(20)^2}=3^2 =9$

Answer 2

Explanation :

(a) Initial velocity, v₁ = 60 mph

Final velocity, v₂ = 20 mph

Let KE₁ and KE₂ are the initial and final kinetic energies.

$\dfrac{KE_1}{KE_2}=\dfrac{1/2mv_1^2}{1/2mv_2^2}$

$\dfrac{KE_1}{KE_2}=\dfrac{(60)^2}{(20)^2}$

$\dfrac{KE_1}{KE_2}=9$

So, the kinetic energy increases 9 times.

(b) Initial velocity, v₁ = 60 mph

Final velocity, v₂ = 30 mph

Let KE₁ and KE₂ are the initial and final kinetic energies.

$\dfrac{KE_1}{KE_2}=\dfrac{1/2mv_1^2}{1/2mv_2^2}$

$\dfrac{KE_1}{KE_2}=\dfrac{(60)^2}{(30)^2}$

$\dfrac{KE_1}{KE_2}=4$

So, the kinetic energy increases 4 times.

Hence, this is the required solution.