Recent content by crimpedupcan

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    Calculating Speed of a Block Pushed Across a Horizontal Surface

    Okay, I've solved the problem using the work/kinetic energy method implied by BruceW. Thanks! If you're curious I got v=\sqrt{\frac{2F_0L}{m}-Lg\mu_0}
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    How to Determine x(t) for a Particle Given Initial Conditions and a(x)?

    Thank you, that's exactly what I was looking for
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    How to Determine x(t) for a Particle Given Initial Conditions and a(x)?

    Thanks a lot! I think I've seen that method applied before to simple harmonic motion, I wish I made the connection earlier.
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    Calculating Speed of a Block Pushed Across a Horizontal Surface

    Homework Statement A block of mass m is at rest at the origin at t=0. It is pushed with constant force F_0 from x=0 to x=L across a horizontal surface whose coefficient of kinetic friction is \mu_k=\mu_0(1-x/L). That is, the coefficient of friction decreases from \mu_0 at x=0 to zero at x=L...
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    How to Determine x(t) for a Particle Given Initial Conditions and a(x)?

    Thanks for the reply. From what I understand your solution requires me to know a(t), but what can I do if I only know a(x)?
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    How to Determine x(t) for a Particle Given Initial Conditions and a(x)?

    Suppose I have a particle on a line, and I know some function a(x) and the initial x, v, and a. How could I work out x(t)?
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    Greatest Moduli Complex Number Solution of Equation

    Okay so nobody commented (maybe I posted it in the wrong place) but I have since found a solution and thought I would post it, in case anyone got curious: Using the triangle inequality: \left|z\right| - \left|\frac{4}{z}\right| ≤ \left|z - \frac{4}{z}\right| ∴ \left|z\right| -...
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    Greatest Moduli Complex Number Solution of Equation

    I would very much appreciate any help with this problem. Homework Statement Find the greatest value of the moduli of the complex numbers z satisfying the equation |z - \frac{4}{z}| = 2 The Attempt at a Solution I tried letting z = a+bi and going from there, but I ended up with this really...
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