Working out frictional force My friend and I did an experiment, and we are to work out the frictional force, but have gone through two different routes with two different answers. I feel like solution "1" is correct, but can't quite figure out why solution "2" isn't correct. Maybe it is! Could you comment please?
                 m2⇓                     m1⇓


The mass $m_1$ falls $(h)$ to the ground and displaces the block $m_2$ to the right by $(h)$. This takes time $t$. The final velocity is calculated using $v=2\frac st$.
To calculate the Force of Friction between $m_2$ and the table, we came up with 
Solution 1
Finding the gravitational potential energy of $m_1$ via $m_1gh$, then deduct this from the kinetic energy of $m_2$, which was calculated via $0.5m_2v^2$.
This gives the work done against friction, and when work done is known, $\frac {work}{distance}=$ Force of Friction
Another way of getting to this figure was finding the weight of $m_1$, thus knowing what force was acting on it, then calculating it's acceleration as it fell via $\frac{2s}{t^2}$. Once the acceleration was known, this was multiplied by $m_1$ to get the Net Downwards Force. This was subtracted from the Weight to find the opposing force, the Force of Friction.
The answers weren't identical, but close.
Solution 2
We know how far $m_2$ displaces, and over what time, so we know it's acceleration. $m_2a$ = Net Force. The difference between the Net Force and applied force (weight of $m_1$) gives the Frictional Force.
I feel like the error in Solution 2 stems from the calculation of the Net Force, by using $m_2$, but can't articulate it well to convince my friend.
What do you think?
 A: Both methods should give the same correct answer, if applied correctly. One uses work done = force x distance, the other uses net force = mass x acceleration.
SOLUTION 1 is correct in theory. However, you have missed out the kinetic energy of mass $m_1$. The PE lost by $m_1$ is equal the the work done against friction plus the KE gained by both masses.
If you use this method you must measure the final velocity of $m_1$ or $m_2$ (they have the same velocity) when $m_1$ has fallen through height $h$. 
SOLUTION 2 seems to be the same method as "another way" mentioned under Solution 1, except applied to $m_2$ instead of $m_1$. The distances moved by $m_2$ and $m_1$ are the same, as are their accelerations, because they are attached by an inextensible string. 
This method is also correct in theory : the net force $m_2$ or $m_1$ equals its mass times its acceleration. However, the applied force is the tension in the string, which is not the weight of $m_1$, it is $m_1(g-a)$ where $a$ is the common acceleration of both masses. If $m_1$ was in free fall $(a=g)$ then the tension in the string would be zero.  
To use this method you must measure acceleration $a$, eg from the time which it takes $m_1$ to fall from rest through height $h$ - or for $m_2$ to move through a distance $h$ from rest. The total force on the 2 masses is $m_1g-F$ where $F$ is the friction force. The total mass being accelerated is $m_1+m_2$. Therefore
$m_1g-F=(m_1+m_2)a$
from which $F$ can be calculated, after $a$ has been measured.
