In the figure, a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed `upsilon_(0)` in `6 m//s`, the height difference h is 1.1 m and the coefficient friction `mu` is 0.6. The value of d is `(g = 10 m//s^(2))`

Explanation

Loss in kinetic energy=Gain in potential energy+Work done against friction
or `1/2mv_0^2=mgh+F_1xxd=mgh+mumgxxd`
Substituting `h=1.1m` and `mu=0.60`, we get
`d=1.2m`