The problem is as follows:
The figure from below shows a set of three pulleys joined by a cable with one end tied to the roof along one of the pulleys. From the center of the first pulley from the right a block of mass "$m$" is hanging as indicated. Meanwhile the other pulley from the right is supported by a stand where a Force $F$ is pulling to the right. It is known that the mass of the block is $2\,kg$ and it takes $3\,s$ to impact to the ground starting from rest. Given these conditions: Find the value of the force $\vec{F}$. (Assume: $g=10\,\frac{m}{s^2}$).
The alternatives given are as follows:
$\begin{array}{ll} 1.&4\,N\\ 2.&6\,N\\ 3.&8\,N\\ 4.&10\,N\\ 5.&12\,N\\ \end{array}$
What i've attempted to do here was to establish this equation.
$F-mg=ma$
But the problem lies on exactly how to define the acceleration in this system. If I go on this route:
$v_f^2=v_o^2+2a\Delta y$
Then
$v_f^2=0+2a\Delta y$
It can be seen from the graph that $\Delta y = 9\,m$
Therefore this can be plug into the previous equation:
$v_f^2=0+2a (9)$
But it also mentions that the time elapsed to hit the ground is $3\,s$. Therefore:
$v_f=v_o+at$
$v_f=0+3a$
If I were to use this into the preceding equation I end up with:
$(3a)^2=0+2a(9)$
And this does produce a contradiction. What am I doing wrong?. Can someone help me here?.
I don't see any contradiction. From the final equation $9a^2=18a$ we can find the solution $a=2$ (for $a\ne0$), that can be used to find $F$.
(But note that the correct equation is $mg-2F=ma$ because the cable is tied to the roof)