Question

A block slides along a track from one level to another level, by moving through intermediate valley. the track is friction less until the block reaches the higher level. there a frictional force stops the block in a distance d. the blocks initial speed $v_0$ is 6 m/s, the height difference h is 1.1 m and the coefficient of kinetic energy $\mu$ is 0.6. the value of d is

Answer 1

$m$ be the mass of the block,

$v_0$ be its initial speed,

$h$ be the height difference,

$d$ be the stopping distance,

$\mu$ be the coefficient of kinetic friction,

$g$ be the acceleration due to gravity.

Equating the block's initial kinetic energy to its increase in potential energy and the work done against friction:

$\frac{m{v}_0^2}{2}=mgh+\mu mgd$

$v_0^2=2g(h+\mu d)$

$h+\mu d=\frac{v_0^2}{\left(2g\right)}$

$d=\frac{[\frac{v_0^2}{\left(2g\right)}-h]}{\mu }$

$d=\frac{[\frac{6^2}{(2\times 10)}-1.1]}{0.6}$

$d=\frac{7}{6}=1.17m$