Question

A body moving at $2$ m/s can be stopped over a distance $x$. lf its kinetic energy is doubled, how long will it go before coming to rest, retarding force remains unchanged?

• A. $x$
• B. $2x$
• C. $4x$
• D. $8x$

Answer 1

Answer: (B)

$2x$

To double the K.E , the new velocity =2$\sqrt{2}m{s}^{-1}$

let the retardation be $=a$

we know $v^2=2\times a\times x$

initial $v=2$

so $2^2=2\times a\times x$ (1)

when $v=2\sqrt{2}m{s}^{-1}$

so$(2\sqrt{2}{)}^2=2a{x}_1-(2)$

From $\left(1\right)and\left(2\right)$, $x_1=2x$