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6 problems on Conservation of Energy

  1. Oct 13, 2007 #1
    I'm in very desperate need of help on one of my homework assignments, I have been getting A's and B's on most of my work, but now I have this. I have been working for several hours without one single answer correct . Any help would be greatly appreciated.
    I went to the professor for help, but he doesn't speak good English and he gets mad when students don't understand physics as well as he does.

    1. The problem statement, all variables and given/known data
    A bead slides without friction around a loop-the-loop as shown in the figure below. The bead is released from a height h = 3.15R.

    <img src="http://www.webassign.net/pse/p8-15.gif"> (if image doesn't work you can copy and past the url in your browser)

    (a) What is the bead's speed (V) at point A? Answer in terms of R and g, the acceleration of gravity.
    (b) How large is the normal force on the bead if its mass is 5.70 g? N (downward)

    1. Relevant equations

    ?

    1. The attempt at a solution

    I know your supposed to attempt a solution before you recieve help (i did on the other problems), but I have no idea of where to go with this, if someone could just point me in the right direction...

    2. The problem statement, all variables and given/known data
    Two objects, m1 = 4.00 kg and m2 = 3.00 kg, are connected by a light string passing over a light frictionless pulley as shown in the figure below. The object of mass 4.00 kg is released from rest, h = 3.50 m above the ground.

    <img src="http://www.webassign.net/pse/p8-13alt.gif">

    (a) Using the isolated system model, determine the speed of the 3.00 kg object just as the 4.00 kg object hits the ground.
    (b) Find the maximum height to which the 3.00 kg object rises.



    2. Relevant equations
    KEi+Ui=KEf+Uf


    2. The attempt at a solution

    KEi+Ui=KEf+Uf
    0+m1gh=1/2m1v^2 +m2gy? (?this solution gives me two variables?)
    plug in numbers and simplify
    137=2v^2 +19.4y
    solve for v
    sqrt(68.5-14.7y)=V
    It seems simply enough, but I don't know how to get around the two variables.

    3. The problem statement, all variables and given/known data
    A daredevil plans to bungee jump from a balloon 76.0 m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test, hanging at rest from a 5.00 m length of the cord, he finds that his body weight stretches it by 1.30 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.
    (a) What length of cord should he use?
    (b) What maximum acceleration will he experience?


    3. Relevant equations
    F=-KX


    3. The attempt at a solution

    Again I have a problem. I am solving both for the skydiver's mass and for K, two separate variables
    F=-KX
    m(9.8)=-K(1.3)?


    4. The problem statement, all variables and given/known data
    A 2.00 kg block is attached to a spring of force constant 455 N/m as in the figure below. The block is pulled 5.85 cm to the right of equilibrium and released from rest.
    (a) Find the speed of the block as it passes through equilibrium if the horizontal surface is frictionless.
    b) Find the speed of the block as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is 0.350.


    4. Relevant equations
    KE=1/2mv^2
    Spring Potential energy: Us=1/2KX
    ?


    4. The attempt at a solution
    K=455
    1/2(455)(.0585)=U
    13.3088=U
    U=KE
    KE=1/2mv^2
    13.3088=1/2mv^2
    13.3088=.5(2)v^2
    13.3088=v^2
    sqrt(13.3088)=V
    3.6481=V
    This is wrong




    5. The problem statement, all variables and given/known data
    A 5.60 kg block is set into motion up an inclined plane with an initial speed of v0 = 8.00 m/s. The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal.

    <img src="http://www.webassign.net/pse/p8-33alt.gif">

    (a) For this motion, determine the change in the block's kinetic energy
    (b) For this motion, determine the change in potential energy of the block-Earth system.
    (c) Determine the frictional force exerted on the block (assumed to be constant).
    (d) What is the coefficient of kinetic friction?


    5. Relevant equations



    5. The attempt at a solution


    I gave up on the last two, sorry

    6. The problem statement, all variables and given/known data
    A 10.0 kg block is released from point A in the figure below. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2300 N/m, and compresses the spring to 0.400 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C.

    <img src="http://www.webassign.net/pse/p8-57.gif">


    6. Relevant equations



    6. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2007 #2
    Problem 1.

    Total energy, E, equals potential energy, V, plus kinetic energy, T, E = V + T.

    Conservation of energy implies total energy is constant, [tex]E_0 = E_A =[/tex] ...

    [tex]E_0 = V_0 + T_0 = mgh + 0 = mgh[/tex]

    [tex]E_A = V_A + T_A = mg\cdot2R + T_A[/tex]

    Now you can solve for [itex]T_A[/itex]

    But you also know that [itex] T_A = v^2/2[/itex]

    Now you can solve for [itex]v[/itex].

    A body moving in a circle is accelerated towards the center of the circle with [itex]a = {v^2} /r[/itex]

    Now you can solve for the force [itex]F_y = -mg + m (-v^2/R)[/itex]

    (In my coordinate system y is directed upward so a negative force is directed downward.)
     
  4. Oct 13, 2007 #3
    Problem 2.
    (a) Conservation implies loss of potential energy equals gain of kinetic energy:

    [tex]m_1gh - m_2gh = (m_1 + m_2)v_{hit}^2/2[/tex]

    Solve for "hit speed" [itex]v_{hit}[/itex].

    (b) Consider only [itex]m_2[/itex]

    [tex]m_2 v_{hit}^2/2 = m_2g (y_{2,max} - h)[/tex]

    Solve for maximum height [itex]y_{2,max}[/itex].
     
  5. Oct 14, 2007 #4
    Problem 3.
    Use these general relations
    Hooke's law F = -(k/L)x
    Gravitation on body F = mg
    Potential energy in cord U = ½*(k/L)x^2
    Potential energy of body U = mgh

    Use Hooke's law and "hanging at rest from a 5.00 m length of the cord, he finds that his body weight stretches it by 1.30 m" to get k of cord.

    Use also that the potential energy his body loses when falling from 76 m to 10 m equals potential energy in string when he momentarily is at rest 10m over the ground.

    Acceleration is maximum when the force is maximum. Force on body is from gravitation (constant) and from cord (variable).
     
  6. Oct 14, 2007 #5
    Thanks Artaxerxes

    I'm working on it right now, I appreciate the help.
     
  7. Oct 15, 2007 #6
    I see that I made a mistake in problem 1. I didn't put the mass into the formula for the kinetic energy.

    Should be [itex]T_A = \frac 12 \cdot m \cdot v^2[/itex].

    :blushing:
     
  8. Oct 15, 2007 #7
    Problem 4.
    (a)
    String potential energy is [itex]U_s =\frac12 \cdot k \cdot x^2[/itex]
    Kinetic energy is [itex]KE = \frac12 \cdot m \cdot v^2[/itex]
    [itex]U_s=KE[/itex] should give you [itex]v[/itex].

    (b)
    Energy lost due to friction is [itex]W_{fr}= F_{fr} \cdot x = \mu \cdot mg \cdot x[/itex]
    The friction force is (almost) independent of speed - that's why you only have to consider the distance.
    [itex]U_s = KE + W_{fr}[/itex] should give you a (lower) value of [itex]v[/itex].
     
  9. Oct 15, 2007 #8
    Problem 5
    (a) For this motion, determine the change in the block's kinetic energy

    Calculate kinetic energy with initial speed v0=8 m/s and then with finel speed v=0. The difference is what they ask for.


    (b) For this motion, determine the change in potential energy of the block-Earth system.

    The change in potential energy is the work done against gravity, mgh. Gravitational force is mg, where m = mass and g is 9,81 m/s2. Only the vertical distance matters here since there is no horizontal component of the gravitational force. You get vertical distance h with a little trigonometry.

    (c) Determine the frictional force exerted on the block (assumed to be constant).

    Frictional force = coefficient of friction times normal force = [itex]\mu \cdot F_n[/itex]
    [itex]F_n = mg \cdot \cos \theta[/itex]


    (d) What is the coefficient of kinetic friction?

    Decrease in kinetic energy = increase in potential energy + loss of energy due to friction

    Loss of energy due to friction = [itex]\mu \cdot mg\cos \theta \cdot x[/itex]
    Now you can solve for [itex]\mu.[/itex]
     
  10. Oct 15, 2007 #9
    Problem 6

    Potential energy at start = energy loss due to friction + potential energy of spring at the moment when the body is at rest

    [itex]mgh = \mu mg L + \frac12 k x^2[/itex]
     
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