Newtons second law and kinematics

[nadinelyab]

1)A rock is dropped off a cliff and falls the first half of the distance into the ground in t1 seconds. If it falls the second half of the distance in t2 seconds, what is the value of t2/t1?

Relevant equations: I think the kinematics equations of d=vt+(1/2)at^2
I have no idea how to solve this please help! I tried to find t1 with the equation above and god t1=sq.rt(d/5) but i don't know if this is right :)

2)a bos of mass m slides on a horizontal surface with initial speed v. I feels no forces other than gravity and the force from the surface. If the coefficient of kinetic friction between the box and the surface is mu how far does the box slide before coming to rest?
Relevant equations: f=ma Ff=Fn*mu,
I basically drew a free body diagram so far

Thanks so much for any help!

 

[azizlwl]

1.
You are using correct kinematics equations of d=vt+(1/2)at^2.
As you see here we have 2 unknowns, d and t with v=0
You need 2 equations to solve for unique solution.

 

[Rokas_P]

Well you're asked for a ratio, t1/t2. Whenever you're asked for a ratio, chances are that something will cancel from both the numerator and the denominator. And I suspect this may just be the unknown distance/height the rock travels, d.

The equation for d is correct as azizlwl pointed out, but since the rock starts from rest, initial speed is zero. So you have d=1/2*at^2.

Okay, you go like this:

1. how long does in take for the rock to fall the first half?

$\dfrac{d}{2}=\dfrac{gt^2}{2}\quad\Rightarrow\quad t=\sqrt{\dfrac{d}{g}}$

and actually this is your t1.

2. how long does it take for the rock to fall down the other half?

You repeat the procedure again but you note that the free fall in the second half of the distance starts with non-zero initial velocity (that's obvious since the rock will have accumulated velocity by the time it is halfway to the ground).

That means the equation for distance traveled is this one now:

$\dfrac{d}{2}=\dfrac{gt^2}{2}+v_0t$

What's $v_0$?

 

[Rokas_P]

OK, since there has been no activity here for quite some time, I'll finish this one off just for the fun of it.

Continuing from my previous post, v₀ is the initial velocity for the second half of the free fall. It is non-zero and is the same as the final velocity the stone will have accumulated by the time it is halfway to the ground. The first half of the free fall started from rest and continued for $t_1=\sqrt{\frac{d}{g}}$ seconds (see my post above). Then the speed accumulated over this many seconds is,

$v_0=gt_1$
$v_0=g\sqrt{\dfrac{d}{g}}=\sqrt{gd}$

Then

$\dfrac{d}{2}=\dfrac{gt^2}{2}+\sqrt{gd}t$

This is a quadratic equation (with respect to t). Its solution gives us the time t₂.

$gt^2+2\sqrt{gd}t-d=0$
$t^2+2\sqrt{\dfrac{d}{g}}t-\dfrac{d}{g}=0$
$t_{1,2}=-\sqrt{\dfrac{d}{g}}\pm\sqrt{2}\sqrt{\dfrac{d}{g}}$

We disregard the negative solution. Then we are left with

$t_2=\sqrt{\dfrac{d}{g}}\left(\sqrt{2}-1\right)$

Ultimately,

$\dfrac{t_2}{t_1}=\sqrt{2}-1$

 

[LawrenceC]

The box problem is one of constant acceleration. Determine the acceleration. Since the problem is one of constant acceleration, it is very similar to an object being projected vertically. How high does it go?

 

[azizlwl]

$\frac {1}{2}d=\frac {1}{2}at{_1}^2...(1)$
$d=\frac {1}{2}a(t_1+t_2)^2...(2)$

from (1)
$d=at_1^2$
sub. in (2)

$at_1^2=\frac {1}{2}a(t_1+t_2)^2$

$2t_1^2=t_1^2+2t_1t_2+t_2^2$

$t_1^2=2t_1t_2+t_2^2$

$1=2\frac {t_2}{t_1} +\frac{t_2^2}{t_1^2}$

$let \ \frac{t_2}{t_1}=x$

$x^2+2x-1=0$

Taking positive root.
x= -1+ √2