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Potential/Kinetic Energies

  1. Sep 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horizontal, by a constant force F with a magnitude of 120N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.25.
    a.) What is the work done on the oven by the force F?
    b.) What is the work on the oven by the friction force?
    c.) Compute the increase in potential energy for the oven.
    d.) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy.
    e.)Use summation{F} = ma to calculate the acceleration of the oven. Assuming the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 14.0m. Compute from this the increase in the oven's kinetic energ, and compare this answer to the one you got in part d.

    2. Relevant equations
    W = F * x
    U = mgy
    K = 1/2(mv^2)
    U1 + K1 + Wother = U2 + K2

    3. The attempt at a solution
    I think there are some typo errors from the back of the book:
    a.) What is the work done on the oven by the force F?
    W = F * x
    W = 120N * (14.0cos(37)) <<<< (x component)
    W = 1341.71 J

    (the answer at back of book is 1690J)

    b.) What is the work on the oven by the friction force?
    Ff = ukn = 0.25(12.0kg)(9.8m/s^2) = 29.4N

    Wf = Ff ( x)
    Wf = (29.4N * 14.0cos(37))
    Wf = 328.72J = 329J
    (the answer at back of book is 329J)

    c.) Compute the increase in potential energy for the oven.
    U2 = mgy2 = (12.0kg)(9.8m/s^2)(14.0sin(37)) = 990.8J = 991J

    (the answer at back of book is 991J)

    d.) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy.

    U1 + K1 + Wother = U2 + K2
    0 + 0 + (WF - Wf ) = U2 + K2
    1341.71 J - 329J - 991J = K2
    K2 = 21.71J

    (the answer at back of book is 360J)

    e.) Use summation{F} = ma to calculate the acceleration of the oven. Assuming the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 14.0m. Compute from this the increase in the oven's kinetic energ, and compare this answer to the one you got in part d.
    F - Ff = ma
    (120N - 29.4N ) / 12.0kg = a
    a = 7.55m/s^2

    vf^2 = v0^2 + 2ax
    vf^2 = 2(7.55m/s)(14.0m)
    Vf = 14.5396m/s

    K = 1/2(mv^2)
    K = 1/2(12.0kg)(14.5396m/s)
    K = 87.238J <<< not same as d i dunno why
    (the answer at back of book is a = 0.699m/s^2: K = 360J)
     
    Last edited: Sep 2, 2007
  2. jcsd
  3. Sep 1, 2007 #2

    bel

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    a) [tex]W= {\int_a}^b F ds [/tex], which is to say you don't only take the x component. (So the book did have a typing error here, but the error is less than one percent.)
    b) The normal force here is not the same as the gravitational force.
    c) [tex] E_K = E_{total} - (E_P + W) [/tex]
    d) Remember that forces are vectors and they point in different directions, so [tex] \sum F [/tex] results in a vector pointing downhill (i.e., direction of acceleration).
     
    Last edited: Sep 1, 2007
  4. Sep 2, 2007 #3
    a.) W = 14.0m(120N)
    W = 1680N

    b.) Ff = uk(mgcos(theta))
    = 0.25(12.0kg(9.8m/s^2)(cos(37))
    = 23.4799N
    Wf = Ff * d
    Wf = 23.4799N(14.0m)
    Wf = 329N

    c.) didnt follow your formula but increase in potential energy:
    U2 = (12.0kg)(9.8m/s^2)(14sin(37))
    U2 = 991J
    same as B.O.B

    d.) U1 + K1 + Wother = U2 + K2
    0 + 0 + (WF - Wf ) = U2 + K2 (U1 and K1 are zero ? because the increase of K2 is required )
    1680 J - 329J = 991J + K2
    K2 = 360J
    yup - same ans at Back

    e.) F - Ff = ma
    (120N - 23.4799N) / 12.0kg = a
    a= 8.043m/s^2
    (not the same as back of book)

    vf^2 = v0^2 + 2ad
    vf^2 = 2( 8.043m/s)(14.0m)
    Vf = 15.004m/s

    K2 = 1/2(mv^2)
    K2 = 1/2(12.0kg( 15.004m/s)^2)
    K2 = 1350.72J
    not also the same

    kindly check this
     
  5. Sep 2, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    a) and b) should be in J... other than that a),b),c) and d) are good.

    For part e) you need to consider the component of gravity acting along the ramp... for the acceleration along the ramp, you need the net force along the ramp...
     
    Last edited: Sep 2, 2007
  6. Sep 2, 2007 #5
    ok
    e.) Fnet = ma
    F - Ff - wsin(theta) = ma
    (120N - 23.4799N - ((9.8*12)(sin(37))/12kg = a
    a = 25.75N / 12kg
    a = 2.1455m/s^2

    vf^2 = 2(2.1455m/s^2)(14.0m)
    vf = 7.75m/s

    K2 = 1/2(12.0kg)(7.75m/s)^2
    K2 = 360.375J

    thanks
     
  7. Sep 2, 2007 #6

    learningphysics

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    Homework Helper

    no prob. It looks good... But you said the back of the book says a=0.699m/s^2 ?
     
  8. Sep 2, 2007 #7
    yes....
    the b.o.b. is 0.699m/s^2
    is that a typo?
     
  9. Sep 2, 2007 #8

    learningphysics

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    Homework Helper

    Yeah, I think so.
     
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