This problem is driving me crazy. It involves an incline, two masses and a rope.

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SUMMARY

The discussion centers on a physics problem involving two masses, m1 = 250g and m2 = 200g, connected by a rope over a frictionless pulley on an incline of 30 degrees. The coefficient of kinetic friction is 0.100. The user struggles with calculating the speed of m2 after it has fallen 30.0 cm, misapplying the tension force and acceleration concepts. The correct approach involves applying Newton's second law to the entire system, recognizing that the tension in the rope is not simply the weight of m2 but must account for the acceleration of both masses.

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This has been driving me crazy because I thought I knew what I was doing, but the book says I'm wrong. I even worked backwards the answer in the book and it makes no sense to me. Please work this out and tell me what I'm doing wrong.

Homework Statement



A block of mass m1 = 250g is at rest on a plane that makes an angle of theta = 30 with the horizontal. The coefficient of kinetic friction between the block and the plane is 0.100. The block is attached to a second block of mass m2 = 200g that hangs freely by a string that passes over a frictionless, massless, pulley. When the second block has fallen 30.0 cm, what will its speed be?

Homework Equations



Fnet = ma (mass is in kg) ; v(final)^2 = v(initial)^2 + 2a(change in x)

kinetic friction = uk Fn ;

The Attempt at a Solution



I started by drawing the first block. I changed the axes so that the surface of the incline would be the x-axis and the y-axis would be the direction of the normal force. So, the forces here should be kinetic friction, tension, normal force, and gravity. I then split gravity (which is .25 * 9.8, or 2.45 N) into its x and y components. Since the incline is 30 degrees, the force of gravity makes a 60 degree angle with its x component. Thus, the x value is 2.45cos(60). And the y value, 2.45sin(60). Thus, the normal force is 2.45sin(60), or 2.12N. I then calculated the magnitude of kinetic friction, which is .212N. I will also mention at this point that the x value of gravity was 1.23 N. Next, I got the force of gravity for the second block, 1.96 N. I think this is the magnitude of the tension of the first block. I then used Newtons second law to get a = 2.07 m/s^2. But I think I must misunderstand tension, because I added the friction force and x value of the gravity force for the first block to get what the tension force on the other block should be. I then used Newtons law again for the second block and got a different acceleration. Either way, I get a different velocity than the one it says in the book. And yes, I used all kilograms and meters.

Summary: I'm going crazy, please help.

EDIT: Why does that template copy when you preview? Whatever.
 
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The mg of the 2nd block is not the tension force. See this by applying
sum of forces = ma to the 2nd block.
mg - T = ma, so T = mg - ma. NOT just mg.

The two blocks and the rope are one system with the same acceleration. Apply
sum of forces = ma
to the whole works. Take that mg as positive, the component of force down the ramp and the friction as negative because they oppose the motion. Use the combined mass in the ma part.

Once you know the acceleration, just use accelerated motion formulas to find the speed.
 
Thanks a lot.
 

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