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Kinetic and Potential Energy

  1. Apr 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Hi, I am taking a physics correspondace course and wondering if I am interpreting these questions correctly. Thanks

    2. Relevant equations
    1. a) A force of 24 N is used to push a smooth box 3.6 m across a smooth floor. What is the final kinetic energy of the box?

    b) IF the box in par a) has a mass of 9.6 kg, calculate its final speed?

    c) A car with a mass of 1.5 X 10^3 kg is travelling along a highway. Calculate its kinetic energy when its speed is
    i) 18 m/s
    ii) 36 m/s

    d) In part c) why is the kinetic energy for a speed of 36 m/s not double the kinetic energy for a speed of 18 m/s?

    2. a) Use an everyday example to explain the terms
    i) potential energy
    ii) gravitational potential energy

    b) A spectator at a hockey game is sitting in a seat situated 10.4 m above the ground level. The ice surface is 2.8 m below ground level. If the spectator has a mass of 52.6 kg, calculate her gravitational potential energy relative to
    i) ground level
    ii) the ice surface

    3. A tobaggan of mass 20.0 kg is dragged horizontally across the snow by a person who exerts a force of 450 N. The toboggan travels 8.0 m under the action of the force and has a final speed of 2.6 m/s.

    a) calculate the actual kinetic energy of the tobaggan based on its speed.

    b) calculate the amount of work actually done on the toboggan by the applied force.

    c) from your answers to parts a) and b) calculate the amount of thermal energy generated .

    d) explain how this thermal energy was generated and state one effect it might have.

    3. The attempt at a solution

    1. a) W = 24 N X 3.6 m
    = 86.4 J
    since w= kinetic energy = 86.4 J

    b) kinetic energy = mv^2 / 2
    v^2 = 2(kinetic energy) / m
    =2 (86.4 J / 9.6 kg
    =18.01 m/s^2
    v = 4.2 m/s

    c) i) Ek = 1500 kg (18m/s)^2 / 2

    ii) at 36 m/s Ek = 1500 mg (36 m/s)^2 / 2
    =972000 J

    d) The kinetic energy for the speed of 36 m/s is actually quadrupled. This is because the value of Ek depends on the square of the speed. Therefore, if the speed increases by a factor of 2 Ek increases by a factor of 2^2 =4

    2. potential energy - the position of objects in the earths gravitational field

    gravitational potential energy - hydropower, such as water in a reservoir behind a dam.

    b) i) ground level - Eg = mg (h)
    =52.6 kg X 9.8 m/s^2 X 10.4 m
    =5360.99 J

    ii) the ice surface - height above ice surface
    h= 10.4 m + 2.8 m = 13.2 m
    Eg = 52.6 kg X 9.8 m/s ^2 X 13.2 m
    =6804.33 J

    3. a) Ek = 1/2 mv^2
    =1/2 (20.0 kg) (2.6 m/s)^2
    =1/2 (20.0 kg) (6.76)
    67.6 J

    b) W = F(d)
    =450 N X 8 m
    =3600 J

    c) Thermal Energy
    The formula is E thermal = E kinetic + E potential
    I have the kinetic energy but do I have to find the formula for potential energy first. I find this confusing because the question says "from your anwers to parts a and b calcualate the amount of thermal energy"
    Not quite sure what I am missing here??????

    d) the amount of friction between the toboggan and the snow causes work to be done as a consequence a transfer of kinetic energy into the molecules on the snow. This increased vibration of the molecules causes an increase in the thermal energy causing the snow to melt.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. 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. Apr 7, 2007 #2


    User Avatar
    Homework Helper

    I think everything looks pretty good. For 3c, look at the total work being done, and the kinetic energy. Think conservation of energy.
  4. Apr 7, 2007 #3
    Thank you.
    E thermal = 67.6 J + 3600 J (amount of work done)
    =3667.6 J
  5. Apr 7, 2007 #4


    User Avatar
    Homework Helper

    Not quite. If 3600 J is the total amount of work done, and after 8 m the energy of the toboggan is 67.6 J, that means that 3600 - 67.6 J of energy must have generated into thermal energy.
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