Question #5b15a

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Answer 1 · 2017-02-12T01:13:52

The kinetic energy increases by a factor of $4$ $4$ .

As you know, the kinetic energy of an object is given by the equation

$color(blue)(ul(color(black)(K_"E" = 1/2 * m* v^2)))$

Here

$m$ is the mass of the object

$v$ is its velocity

This equation shows that the kinetic energy of an object varies with the square of its velocity.

Let's say that you have

$K_"E 1" = 1/2 * m* v_1^2$

If you double the object's velocity

$v_2 = 2 * v_1$

then its kinetic energy will increase by a factor of $4$ , since

$K_"E 2" = 1/2 * m * v_2^2$

$K_"E 2" = 1/2 * m * (2 * v_1)^2$

$K_"E 2" = 1/2 * m * v_1^2 * 2^2$

But since

$K_"E 1" = 1/2 * m * v_1^2$

you can say that

$K_"E 2" = K_"E 1" * 2^2$

which is equivalent to

$color(darkgreen)(ul(color(black)(K_"E 2" = 4 * K_"E 1")))$