Recent content by Aggression200

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    Finding final velocity.

    Determine the kinematic equation that you would have to use..... Vf = Vi + at Delta x = .5(vf+vi)t Delta x = Vi(t)+.5a(t^2) Vf^2 = Vi^2 + 2a(delta x) Choose which one you would use, and solve for Vf.
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    Acceleration of the block

    Consider that the Force being applied is someone pulling on a string. That string is 60.0 degrees to the flat surface the block is on. Ergo, the Force is at 60.0 degrees to the horizontal. And the friction is acting on the flat surface, keeping the angle at 180. But, to answer your question...
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    Acceleration of the block

    You wouldn't multiply the frictional force by cos(theta) though. If you draw your FBD.... You have a friction force going straight backwards with a magnitude of 20 N. Therefore, it has an angle of 180 with direction of motion, so cos(180) = -1. The 60.0 degrees ONLY affects the diagonal force...
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    Distance From a Velocity-Time Graph

    You know that Velocity is a change in position over change in time, correct? Using that, you should be able to figure out the position. (At least, that's how I do it.)
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    Acceleration of the block

    You have to remember that the friction force also applies here. \SigmaF = m*a \SigmaF = Fcos(theta) - friction = ma Correct?
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    Two Masses Connected by Pulley; find acceleration

    In order to come up with these equations, you have to first figure out what forces are acting upon the objects in question. The four basic ones when dealing with pulleys are: 1. Weight (otherwise known as gravitational) 2. Normal (otherwise known as natural in some cases) 3. Tension 4...
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    Two Masses Connected by Pulley; find acceleration

    If you want, I can try to explain how to get these equations. :smile:
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    Two Masses Connected by Pulley; find acceleration

    Now, you add your 2 equations together..... \SigmaF = -m_{2}g - \mum_{1}g = m_{1}a + m_{2}a Solve for a.
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    Two Masses Connected by Pulley; find acceleration

    M1: Normal (upward), Weight (downward), Tension (rightward), Friction (rightward) M2: Tension (upward), Weight (downward) M1's equation \SigmaF = m_{1}g - m_{1}g - T_{1,2} - \mum_{1}g = m_{1}a M2's equation \SigmaF = -m_{2}g + T_{2,1}= m_{2}a Correct?
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    3 Sliding Masses; find speed

    m2g canceled out because the equation for M2 included..... m2g - m2g = 0 So, that's why they canceled out.
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    3 Sliding Masses; find speed

    Let's assume they are. haha Now, add up all the equations. The tension forces will cancel leaving you with..... \SigmaF = m_{3}g - m_{1}g = m_{1}a + m_{2}a + m_{3}a So, factor out a g from the left side, and an a from the right side..... g(m_{3}-m_{1}) = a(m_{1} + m_{2} + m_{3})...
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    3 Sliding Masses; find speed

    Well, after you draw your FBDs, you will have a system of 3 equations. So, sum the forces in each situation. We will denote T_{1,2} as the tension force between the M1 and M2 object. We will denote T_{2,3} as the tension force between the M2 and M3 object. T_{2,1} = T_{1,2} T_{3,2} =...
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    Block sliding Down; find kinetic coefficient

    No matter how many times I do this problem, I still get .305 as uk..... So, I don't know. Maybe there was a rounding thing that was done somewhere to make it different than .305.
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    Block sliding Down; find kinetic coefficient

    I'm sorry. I'm honestly stuck on this problem. I think I summed the forces right (since I could solve for us and get the right answer). But, when it gets to finding the acceleration, I'm confused. Maybe if I think about it some more.....
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    3 Sliding Masses; find speed

    Alright, now we can get this problem started. So, you'll have 3 separate FBDs. M1's FBD will have a T force going up and a Weight Force going down. M2's FBD will have tension forces on both sides, a normal force going up that is equal to a weight force going down. M3's FBD will have a T...
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