Two block on each other with friction

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The discussion revolves around calculating the forces acting on two blocks with friction. The initial calculations for the friction force and normal force are questioned, particularly whether the correct normal force was used in the equations. There is a consensus that the normal force should account for both weights, as the net vertical force is zero. The friction force acting on the first mass needs clarification regarding its source, whether from the second mass or the ground. The participants aim to identify the mistake in the calculations to arrive at the correct answer.
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Homework Statement


screenshot_24.png


Homework Equations


F = ma
Ffr = u * Fn

The Attempt at a Solution


After drawing FBD
For m2: we have only friction force in horizontal axis so Ffr = ma ==> u * Fn = ma ==> u * mg = ma ==> u * g = a ==> a = 0.5 * 10 = 5 m/s2
Now, moving to m1:
F - Ffr = ma ==> F - u Fn = 10 * 5 ==> F - 0.5*130 = 50 ==> F - 65 = 50 ==> F = 115 N

But it is not in choices,

Is my solution wrong?
or
Are the choices wrong?
 
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One of the choices is right. Your solution is wrong.

a = Fnet/m

What mass is needed in this equation?
 
Sarah00 said:
Now, moving to m1:
F - Ffr = ma ==> F - u Fn = 10 * 5 ==> F - 0.5*130 = 50

Did you use the correct value for the normal force here?
 
I though that the mistake is with the Fn.
Fn here is the sum of two weights as net vertical force is zero

Dr Courtney, you considered them as single mass I see. The answer will be 65 N then. but I want to know my mistake
 
Sarah00 said:
I though that the mistake is with the Fn.
Fn here is the sum of two weights as net vertical force is zero

Think about whether the friction force acting on m1 is coming from m2 or coming from the ground.

What is the normal force acting between m1 and m2?
 
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