A track has two inclined surface AB and DC each of length 3 m and angle of inclination of `30^@` with the horizontal and a central horizontal part of length 4m shown in figure. A block of mass 0.2 kg slides from rest from point A. The inclined surfaces are frictionless. If the coefficient of friction between the block and the horizontal flat surface is 0.2, where will the block finally come to rest ? [in `10^(-1) m`]

Answer

`0.5`

Explanation

At point A, the energy of the block is entirely potential = mgh, where h is the height of A above the horizontal surface = AB `sin theta = 3 m xx sin 30^@ = 1.5 m` . When the block is released, it reaches the point B where the entire potential energy is converted into kinetic energy. It will move on track BC. On reaching C, it will rise on the surface CD to a point E and will return to C moving towards B and the process goes on until the block loses all its energy and comes to rest. The body loses energy while moving on the horizontal part BC of the track. If s is the total distance travelled on the horizontal part, the work done against friction is `W = mu R xx s = mu m gs`
From the principle of conservation energy , we have mgh = `mu m gs`
`s = h/(mu) = (1.5 m)/(0.2) = 7.5 m`.
i.e., the block travels a total distance of 7.5 m on the horizontal track BC before coming to rest, starting from B. It travels 4m from B to C and 3.5 m from C to a point P, where it comes to rest. Hence, the block comes to rest at the point P at a distance of 0.5 m from B.