Calculating Acceleration and Tension in a Two-Block System with Friction

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To calculate the acceleration of the two blocks, the net force must account for friction, which is calculated using the coefficient of friction (µ = 0.23). The correct approach involves determining the frictional force (Fk) as Fk = µ * (m1 + m2) * g, where g is the acceleration due to gravity. After finding the frictional force, the net force can be recalculated as F_net = Fp - Fk, leading to the correct acceleration using F_net = (m1 + m2) * a. Once the acceleration is accurately determined, the tension in the string can be calculated using T = m2 * a, where m2 is the mass of the second block. Understanding the role of friction is crucial for solving both parts of the problem accurately.
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A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of Fp = 30 N. The coefficient of friction of the blocks with the floor is µ = 0.23.

a) What is the acceleration of the two blocks?

For this question, I used Newton's Second Law. I used F=ma. I did 30=(5 kg)(a). I got 6 m/s2 for the acceleration, but that's not right. How do I do the acceleration then?

b) What is the tension in the string between the blocks?
I think I can do b if I get the accelration from part a. I think the equation is T=ma.
 
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For a, you're forgetting friction in the net force sum. Try b after you get a.
 
Is that Fk=(mew)(mass)(acceleration)? But then what's Fk?
 
I think you're misunderstanding what exactly the coefficient of kinetic friction is. It is the ratio of the frictional force to the normal force of the object.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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