Forces/Dynamics: A block sliding on a plate

[haruspex]

So something like:
$a = \frac {(F - (m_b*a_b))}{m_p}$
$a = \frac {(2TN - (m_b*a_b))}{m_p}$
?

Why do you still have a_b terms on the right? There is only the one acceleration during this phase of the problem. Turn all a_p and a_b terms into plain a and rearrange to get the equation into the form a=some function of F, m_b and m_p. There should be no a terms on the right.
When you have that, substitute 2t for F and see if you can get an equation for the velocity.

 

[jisbon]

Why do you still have a_b terms on the right? There is only the one acceleration during this phase of the problem. Turn all a_p and a_b terms into plain a and rearrange to get the equation into the form a=some function of F, m_b and m_p. There should be no a terms on the right.
When you have that, substitute 2t for F and see if you can get an equation for the velocity.

Oh sorry. Corrected it as followed:
$a = \frac {(F - (m_b*a))}{m_p}$
$a*m_p = (F - (m_b*a))$
$(a*m_p) + (m_b*a) = F$
$a(m_p + m_b) = F$
$a = \frac {F}{(m_p + m_b)}$
$a = \frac {2tN}{(m_p + m_b)}$
To be clear, I will be using N for the plate right?
This a I'm finding will be the acceleration before it slips. So from there, I am able to find the velocity of plate when it starts to slip with this acceleration value? Thanks!

 

[haruspex]

from there, I am able to find the velocity of plate when it starts to slip with this acceleration value?

I would call it an acceleration function, not a value; it is not constant.
But yes, you now have the acceleration as a function of time. How do you find the velocity as a function of time from that?

 

[jisbon]

I would call it an acceleration function, not a value; it is not constant.
But yes, you now have the acceleration as a function of time. How do you find the velocity as a function of time from that?

I will integrate it to become velocity?
So from my previous equation:

$a=\frac {2tN}{m_p+m_b}$

$a = \frac {2t*9.8*1.5}{1.5+0.5}$
$a = 14.7t$
$v = (\frac {14.7}{2})t^2$?

 

[haruspex]

I will integrate it to become velocity?
So from my previous equation:

$a = \frac {2t*9.8*1.5}{1.5+0.5}$
$a = 14.7t$
$v = (\frac {14.7}{2})t^2$?

Yes, your integration is fine, but where did the factor g come from?

 

[jisbon]

Yes, your integration is fine, but where did the factor g come from?

Oh wait isn't $N= g* mass$?

 

[haruspex]

Oh wait isn't $N= g* mass$?

No. N stands for Newtons. A Newton is the force required to accelerate a mass of 1kg at 1m/s². Nothing to do with gravity.

 

[jisbon]

No. N stands for Newtons. A Newton is the force required to accelerate a mass of 1kg at 1m/s². Nothing to do with gravity.

Oh no. I always thought that the formula was F=2tN where N is the normal force, turns out N is supposed to be the units :/
If that's the case will $a= \frac {2t}{1.5+0.5} = t$
Hence $v = \frac {1}{2}t^2$
?

 

[haruspex]

Oh no. I always thought that the formula was F=2tN where N is the normal force, turns out N is supposed to be the units :/
If that's the case will $a= \frac {2t}{1.5+0.5} = t$
Hence $v = \frac {1}{2}t^2$
?

Yes,

 

[jisbon]

So with a and v, I can find out initial velocity when plate before it slips?

 

[haruspex]

So with a and v, I can find out initial velocity when plate before it slips?

Yes - do you see how?