If the work done to double the velocity of a body from \(15\ \mathrm{ms^{-1}}\) is \(K\) times the work done to double its velocity from \(10\ \mathrm{ms^{-1}}\), then the value of \(K\) is
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If a body's speed is doubled,
\[
\boxed{
W=\Delta KE
=\frac12m\left((2u)^2-u^2\right)
=\frac32mu^2.
}
\]
Thus, the work required is directly proportional to the square of the initial speed.