What happens to the kinetic energy of an object if we halve the velocity?

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What happens to the kinetic energy of an object if we halve the velocity?

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Answer 1 · 2017-12-16T17:15:10

See below.

Explanation:

Kinetic energy is given by $K = \frac{1}{2} m v^{2}$ $K=1/2mv^2$

where $m$ $m$ is the mass of the object, and $v$ $v$ is the object's velocity (speed).

If we halve the object's velocity, we're saying that the final velocity is $\frac{1}{2}$ $1/2$ the initial velocity, or in symbolic terms, $v_{f} = \frac{1}{2} v_{i}$ $v_f=1/2v_i$ .

So, initially we would have $K_{i} = \frac{1}{2} m v_{i}^{2}$ $K_i=1/2mv_i^2$ and finally we would have:

$K_{f} = \frac{1}{2} m v_{f}^{2}$ $K_f=1/2mv_f^2$

$= \frac{1}{2} m \cdot {(\frac{1}{2} v_{i})}_{i}^{2}$ $=1/2m*(1/2v_i)^2$

Note that the entire quantity—including both the $\frac{1}{2}$ $1/2$ and $v$ $v$ term—is squared.

$\Rightarrow \frac{1}{2} m \cdot \frac{1}{4} v_{i}^{2}$ $=>1/2m*1/4v_i^2$

Rearranging:

$\Rightarrow \frac{1}{4} (\frac{1}{2} m v_{i}^{2})$ $=>1/4(1/2mv_i^2)$

$= \frac{1}{4} K_{i}$ $=1/4K_i$

$:.$ We can see that the final kinetic energy of the object is $1/4$ of the kinetic energy initially possessed by the object.

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What happens to the kinetic energy of an object if we halve the velocity?

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