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    First one is a Work Done / Power question.

    If the air resistance to the motion of an airliner at speed vms^{-1} is given by kv^2 newtons at ground level, then at 6000 meters the corresponding formula is 0.55kv^2, and at 12000 meters it is 0.3kv^2. If an airliner can cruise at 220ms^{-1} at 12000, at what speed will it travel at 6000 meters with the same power output from the engines.

    I've done this bit, the answer being 180ms^{-1}. It is the second part I cannot do:

    Suppose that k=2.5 and that the mas of the airliner is 250 tonnes. As the airliner takes off its speed is 80ms^{-1} and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?


    --------------------------------


    Second is this Potential Energy question:

    Two particles of mass 0.1 kg and 0.2 kg are attached to the ends of a light inextensible string which passes over a smooth peg. Given that the particles move vertically after being released from rest, calculate their common speed after each had traveled 0.6 meters.

    Done this bit, answer being 1.98ms^{-1}. It is the next two bits I cannot grasp:

    Deduce the work done on the lighter particle by the string, and use this to calculate the tension in the string.

    --------------------------------

    I would really appreciate an explanation of the second one especially, including an in-depth description of included forces and why we use whatever you used to get the answers.

    Thanks for everyone' time.
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    (Original post by SamKeene)
    First one is a Work Done / Power question.

    If the air resistance to the motion of an airliner at speed vms^{-1} is given by kv^2 newtons at ground level, then at 6000 meters the corresponding formula is 0.55kv^2, and at 12000 meters it is 0.3kv^2. If an airliner can cruise at 220ms^{-1} at 12000, at what speed will it travel at 6000 meters with the same power output from the engines.

    I've done this bit, the answer being 180ms^{-1}. It is the second part I cannot do:

    Suppose that k=2.5 and that the mas of the airliner is 250 tonnes. As the airliner takes off its speed is 80ms^{-1} and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?


    --------------------------------


    Second is this Potential Energy question:

    Two particles of mass 0.1 kg and 0.2 kg are attached to the ends of a light inextensible string which passes over a smooth peg. Given that the particles move vertically after being released from rest, calculate their common speed after each had traveled 0.6 meters.

    Done this bit, answer being 1.98ms^{-1}. It is the next two bits I cannot grasp:

    Deduce the work done on the lighter particle by the string, and use this to calculate the tension in the string.

    --------------------------------

    I would really appreciate an explanation of the second one especially, including an in-depth description of included forces and why we use whatever you used to get the answers.

    Thanks for everyone' time.
    Hi Sam,

    Question 2 first:

    You can see that lighter particle has travelled 0.6m up and it's speed has increased from 0 to 1.98.

    This corresponds to an increase in gravitational potential energy (since the particle has been raised from its starting position) and also an increase in kinetic energy (since the particle's speed has increased)

    This increase in energy has got to come from somewhere, It comes from the work done on the particle by something external to it. In this case the rope.

    So to calculate the work done just calculate the total increase in the particle's energy.

    Now work done is itself equal to force x distance moved.

    The resultant force on the particle is T-0.1g and the distance moved is 0.6 metres. Now just solve for T.

    Hope this helps.
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    (Original post by SamKeene)
    First one is a Work Done / Power question.

    If the air resistance to the motion of an airliner at speed vms^{-1} is given by kv^2 newtons at ground level, then at 6000 meters the corresponding formula is 0.55kv^2, and at 12000 meters it is 0.3kv^2. If an airliner can cruise at 220ms^{-1} at 12000, at what speed will it travel at 6000 meters with the same power output from the engines.

    I've done this bit, the answer being 180ms^{-1}. It is the second part I cannot do:

    Suppose that k=2.5 and that the mas of the airliner is 250 tonnes. As the airliner takes off its speed is 80ms^{-1} and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?


    --------------------------------


    Second is this Potential Energy question:

    Two particles of mass 0.1 kg and 0.2 kg are attached to the ends of a light inextensible string which passes over a smooth peg. Given that the particles move vertically after being released from rest, calculate their common speed after each had traveled 0.6 meters.

    Done this bit, answer being 1.98ms^{-1}. It is the next two bits I cannot grasp:

    Deduce the work done on the lighter particle by the string, and use this to calculate the tension in the string.

    --------------------------------

    I would really appreciate an explanation of the second one especially, including an in-depth description of included forces and why we use whatever you used to get the answers.

    Thanks for everyone' time.
    For q1a is the speed 180 ms per sec?

    I get the following:

    At ground level the air resistance is kv^2
    At 6000m the air resistance is 0.55kv^2
    At 12000m the air resistance is 0.3kv^2

    When at 12000m and travelling at 220 m/s the air resistance is
    0.3*k*220^2 = 14520k

    Since the plane is cruising there is no resultant force so the engines thrust = air resistance

    So engines thrust = 14520k

    Now at 6000m we are told the plane is cruising again. And we are told that the thrust is the same.

    So:

    Resultant force = 0
    Therefore Engines thrust = Air resistance
    14520k = 0.55kv^2
    14520/0.55 = v^2
    v = 162.48 m/s
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    (Original post by ApplyYourself)
    For q1a is the speed 180 ms per sec?

    I get the following:

    At ground level the air resistance is kv^2
    At 6000m the air resistance is 0.55kv^2
    At 12000m the air resistance is 0.3kv^2

    When at 12000m and travelling at 220 m/s the air resistance is
    0.3*k*220^2 = 14520k

    Since the plane is cruising there is no resultant force so the engines thrust = air resistance

    So engines thrust = 14520k

    Now at 6000m we are told the plane is cruising again. And we are told that the thrust is the same.

    So:

    Resultant force = 0
    Therefore Engines thrust = Air resistance
    14520k = 0.55kv^2
    14520/0.55 = v^2
    v = 162.48 m/s
    The book gave the answer 180. I got the same answer by:

    Power = Fv (Force x Velocity)
    Engine thrust = 14520K (same way you got it)

    Power = Fv = 14520k * 220 = 3194400k

    Same power output at 6000m

    F=0.55kv^2

    Fv=0.55kv^3

    Power = 0.55kv^3

    So 3194400k=0.55kv^3

    Cancel the K's and solve for v.

    V= 179.7527...

    I think your different answer is that you've done it having the same thrust from the engines, not power (which would be F*v = Thrust*220)
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    (Original post by SamKeene)
    The book gave the answer 180. I got the same answer by:

    Power = Fv (Force x Velocity)
    Engine thrust = 14520K (same way you got it)

    Power = Fv = 14520k * 220 = 3194400k

    Same power output at 6000m

    F=0.55kv^2

    Fv=0.55kv^3

    Power = 0.55kv^3

    So 3194400k=0.55kv^3

    Cancel the K's and solve for v.

    V= 179.7527...

    I think your different answer is that you've done it having the same thrust from the engines, not power (which would be F*v = Thrust*220)
    I see the error of my ways :rolleyes:

    I'll still try part b though.

    For q2 is T=2.2887N ?
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    (Original post by ApplyYourself)
    I see the error of my ways :rolleyes:

    I'll still try part b though.

    For q2 is T=2.2887N ?
    The answer from the book is 0.784 J for the work done - and 1.31 N for the tension.

    Damn these tricky mechanics questions!
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    (Original post by SamKeene)
    The answer from the book is 0.784 J for the work done - and 1.31 N for the tension.

    Damn these tricky mechanics questions!
    I retire lol.

    Are they M2 questions?

    I'm going to have lunch, refocus and come back here with answers (the right answers)
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    (Original post by ApplyYourself)
    I retire lol.

    Are they M2 questions?

    I'm going to have lunch, refocus and come back here with answers (the right answers)
    Yeah M2.

    I'll probably give it another go today as well at some point.
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    (Original post by SamKeene)
    The answer from the book is 0.784 J for the work done - and 1.31 N for the tension.

    Damn these tricky mechanics questions!
    I think I'm there with one of the questions.. I've attached my answer.
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