Question

(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 rest through 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.

Answer 1

a. Let a particle sliding from C to A along the inclined plane CA acquire a final velocity $v_1$, covering a distance s.
If $\Theta$ is the angle of inclination, 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, $v^2=u^2+2as$
$\Rightarrow v_1^2=2\left(gsin\Theta \right)s=2g\left[h/s\right]s$ = 2gh
$\therefore v_1=\sqrt{2gh}$ ...(i)
Again, let a particle fall from rest along a distance h and acquire a velocity $v_2$, then from $v^2=u^2+2as$,
$v_2^2=0+2gh\Rightarrow v_2=\sqrt{2gh}$ ...(ii)
From (i) and (ii), we get the same result.
b. Time taken 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)(s sin$\Theta$)$t^2$
$\Rightarrow t^2=\frac{2s}{ssin\Theta }=\frac{2h}{gsi{n}^2\Theta }$
$\therefore t=\sqrt{\frac{2h}{g}}$cosec$\Theta$
Thus, the time taken varies directly as the cosecant of the angle of inclination. Now, since cosecant is a decreasing function in the first quadrant with an increase in $\Theta$, time t would decrease.