(a) Show that the velocity acquired by a particle in sliding down an inclined plane is the same as that acquired by a particle falling freely from frst though a distance equal to the height of the inclined plane. (b) Find the time taken in sliding a particle down the whole length of the incline.

Explanation

Let a particle sliding from `c` to `A` along the inclined plance`CA` acquire a final velocity `v_(1)`, comering a distance `s` .
If `theta` is the angle of inlination, then
`sin theta =(h)/(s)`
.
Now for the sliding particle, `s=s,u=0,a=g sin theta, v=v_(1)`.
[Taking the direction `C` to `A` as positive]
Using, `v^(2)u^(2)+2as`
`rArr v_(1)^(2) =2(g sin rheta)s =2g[(h)/(s)]s=2 gh`
`becuase v_(1) =sqrt2gh`
Again, let aparticle fall from rest along a distance `h` and acquire a velocity `v_(2)`, then from `v^(2)=u^(2)+2as`,
`v_(2)^(2)=0+2 gh rArr v_(2)=sqrt 2gh` (ii)`
From (i) and (ii), we get the same result.
b. Time take in sliding a particle down the entire length of the incline:
For the motion of particle from C to A `u=0`,
`s=s, a=g sin theta, t=`?.
Using `s=ut+(1)/(2) a t^(2)`, we get `s=(1)/(2)(g sin theta) t^(2)`
`rArr (t^(2)=(2s)/(g sin theta)=(2h)/(g sin^(2) theta)` [becuase sin theta=(h)/(s)]
`becuase t= sqrt(2h)/(g) cos theta`
Thus, the time take varies directly ad the cosecant of the angle of incliation. Now, since cosecant is a dectrasing function in the first quadrant with an increase in `theta` time `t` would decrease.