How long does it take this object to arrive at the point $M$? Assume that $|KL| =|LM|$.

Question

• How long does it take this object to arrive at the point $M$? Assume that $|KL| =|LM|$.

Options:

a) $\frac{3}{2}\sqrt \frac{h}{g\sin^2\theta}$ b) $\frac{3}{2}\sqrt \frac{2h}{g\sin^2\theta}$ c) $\frac{3}{2}\sqrt \frac{2h}{g\cos^2\theta}$ d) $\frac{3}{2}\sqrt {2h} \sin \theta$ e) $\frac{3}{2}\sqrt {2h} \cos^2 \theta$

I'm out of my mind right now. It seems very complex to me, doesn't it? In fact, I've to be familiar with free-body diagram to solve this question easily. Can I take your professional helps?

Regards!

Answer 1

Use Newton's equations of motion

$mg\sinθ=ma$

$⇒a=g\sinθ$

$v^2=2g\sinθ*h/\sinθ$ (velocity on point L).

$v=g\sinθt_1$

$t_1=\dfrac{\sqrt{2gh}}{g\sinθ}$

$t=t_1+t_2=\dfrac{\sqrt{2gh}}{g\sinθ}+\dfrac{h}{\sqrt{2gh}\sinθ} =\dfrac{3\sqrt h}{\sqrt{2g}\sinθ}$

this solution holds on difficult situation which square-like object continue to slide with vertex from point L.

Answer 2

1)$ |KL|:=s.$

Note : $h/s= \sin \theta$; or

$s= h/(\sin \theta)$.

Constant $a=g \sin\theta$ down the slope.

A) Final speed at $L$:

$v_f = at$ , where $t$ is the time taken down the slope, and

$s= (1/2)at^2$, or

$t_1:= t =((2s)/a)^{1/2}$; and

$v_f = a((2s)/a)^{1/2}$.

2) Constant speed $v_f$ along $|LM|=s$ ;

$t_2:= t= s/v_f.$

3) Total time taken from $K$ to $L$:

$T= t_1+t_2$.