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Relationship between force and (kinetic) energy

  1. Nov 20, 2014 #1
    A car of mass 1kg rests on top of a large flat mass of 1,000,000 kg (10^6 kg). They are at rest (relative to the sun) in space. The car has a battery and an electric motor with 100% efficiency. We can assume that when a torque is applied to the wheels, they roll on the surface of the flat mass without any slippage or rolling resistance.

    The engine applied a torque to the wheel for 1 second in such a way that the force on the car is exactly 1N.

    m1 = 1 kg, m2 = 10^6 kg, dt = 1s, F = 1 N.


    Using F = ma, we get:
    F = m1 * a1;
    a1 = F/m1 = 1/1;
    a1 = 1 N/s^2 (acceleration of car). Since dt = 1 s, we get:
    v1 = 1 m/s;

    Similarly, we find that that
    V2 = -0.000001 m/s = -10^-6 m/s.

    The principle of conservation of momentum gives the same result.
    Now the car is travelling along the flat mass with a velocity of 1m/s (relative to the sun) and a relative velocity of 1.000001m/s relative to the flat mass.

    The kinetic energy of the car relative to the sun is KE1 = ½ * m1*v1^2.
    KE1 = ½*1*1*1 = 0.5 J.

    The kinetic energy of the flat relative to the sun is KE2 = ½ * m2*v2^2.
    KE1 = ½*10^6*10^-6*10^-6 = 5*10^-7 J.

    Total KE is 0.5000005 J.

    All this energy came from the battery of the car and all of it got converted to kinetic energy.

    At this point the engine applied another torque to the wheel for 1 second second in such a way that the force on the car is exactly 1N.

    Can we assume that the amount of energy required if almost the same for the second period of applied force as it was for the first? Why?
     
  2. jcsd
  3. Nov 20, 2014 #2

    A.T.

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    What does the definition of work say about applying the same force, for the same duration, but at a higher speed?
     
  4. Nov 20, 2014 #3
    Since the work done on a body is equal to the force times distance covered when the force is being applied. So in this case more work will have to done to apply the same force for the same duration. Meaning more energy is required to achieve the same amount of acceleartion.

    But lets look at the actual power transmission from the battery to the engine to the wheels. When the car is moving at constant velocity and there are no power loses, the engine does not apply any torque. Assuming that the same amount of power is flowing from the battery to the engine, am I right to assume that the torque transferred to the wheels will be less than in the first case?
     
  5. Nov 20, 2014 #4

    A.T.

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    It's not clear to me what cases you are comparing.
     
  6. Nov 20, 2014 #5
    Case 1: Car is at rest and engine delivers torque for one second.

    Case 2: Car is moving at a constant velocity on the flat mass and the engine delivers a torque using the same amount of energy as in case 1.
     
  7. Nov 20, 2014 #6

    A.T.

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    Didn't you just conclude yourself, that assuming no loses, there won't be any torque in this case?
     
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