Question #19fce

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Question #19fce

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Answer 1 · 2017-03-27T18:25:41

Let the block of mass $m$ $m$ be projected along the smooth inclined plane having angle of inclination $α$ $alpha$ with the horizontal and it moves distance $s$ $s$ along the inclined plane in time t. During this time if it achieves a vertical height h,then $h = s sin α$ $h=ssinalpha$

Now retardation of the block along the inclined plane is $g sin α$ $gsinalpha$

So $s = v t - \frac{1}{2} g sin α t^{2}$ $s=vt-1/2gsinalphat^2$

And

$h = s sin α$ $h=ssinalpha$

$\Rightarrow h = v sin α t - \frac{1}{2} g {sin}^{2} α t^{2}$ $=>h=vsinalphat-1/2gsin^2alphat^2$

So potential enrgy of the block at time t will be

$E_{p} (t) = m g h$ $E_p(t)=mgh$

$= v m g sin α t - \frac{1}{2} m g^{2} {sin}^{2} α t^{2}$ $=vmgsinalphat-1/2mg^2sin^2alphat^2$

$= a t - b t^{2}$ $=at-bt^2$

Where $a and b$ $a and b$ are constants as $v, m, g and α$ $v,m,g and alpha$ are constants.
So a plot of $E_{p} (t)$ $E_p(t)$ v/s $t$ $t$ will be a parabola for the upward and downward movement of the block along the inclined plane.

So option (1) will be the answer.

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Question #19fce

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