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anonymous

  • 4 years ago

PHYSICS QUESTION. helpppp with this last problem.

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  1. anonymous
    • 4 years ago
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    An object's mass is 50 KG. 1- what is its potential energy relative to the ground if its height is 20 meters above the earth? 2- What is its potential energy relative to the ground if the objet is raised from 20m to 100 m? 3- what is the gain in potential energy? 4- What is the gain in total mechanical energy ? (assuming the object at rest at both 20 and 100 m)

  2. anonymous
    • 4 years ago
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    5- What is the amount of work applied to the object to raise it from 20m to 100m? 6- what is the amount of force applied to raise it? 7- what is the weight of the object? 8 - why is the weight equal the force acting on the object lift it? 9 - what is the power of the machine used to do lifting if the process took 5s to complete?

  3. anonymous
    • 4 years ago
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    What is the equation for potential energy?

  4. anonymous
    • 4 years ago
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    m * g * y

  5. anonymous
    • 4 years ago
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    Correct. So what is the potential energy when y = 20?

  6. anonymous
    • 4 years ago
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    So 1 = 9800 J and 2 = 39200J but with 3 i subract 39200 - 9800?

  7. anonymous
    • 4 years ago
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    Yes, yes, and yes!

  8. JamesJ
    • 4 years ago
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    @nikkysterr: don't forget eashmore some medals for his help!

  9. anonymous
    • 4 years ago
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    whooo ! alright but it confuses me when it comes to "object at rest"

  10. anonymous
    • 4 years ago
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    For #4, total mechanical energy is the sum of potential and kinetic energies. If an object is at rest, what is it's kinetic energy?

  11. anonymous
    • 4 years ago
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    0?

  12. anonymous
    • 4 years ago
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    Correct. So what is the total mechanical energy?

  13. anonymous
    • 4 years ago
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    Mechanical Energy is zero.

  14. anonymous
    • 4 years ago
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    I'm having some internet trouble. Forgive me if my responses are delayed.

  15. anonymous
    • 4 years ago
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    no problem i appreciate your help.

  16. anonymous
    • 4 years ago
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    physics isnt my strongest subject.

  17. anonymous
    • 4 years ago
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    Nope. Mechanical Energy (ME) is defined as such\[ME = KE + PE\]If KE is zero, we still need to account for PE.

  18. anonymous
    • 4 years ago
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    so mechanical energy isnt zero?

  19. anonymous
    • 4 years ago
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    Nope. It is equal to potential energy in this case.

  20. anonymous
    • 4 years ago
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    Oh so the mechanical energy would be the Potential Energy?

  21. anonymous
    • 4 years ago
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    for the amount of work how do i find the force?

  22. anonymous
    • 4 years ago
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    i know i have the distance 80 [100-20]

  23. anonymous
    • 4 years ago
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    but i need the force. to get the work.

  24. anonymous
    • 4 years ago
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    Nope. To find the work, we can use the work-energy theorem which is defined as\[W = \Delta ME = \Delta KE + \Delta PE = \Delta PE\]

  25. anonymous
    • 4 years ago
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    We will find the force from this value of work, using the definition of work you are thinking of. You're on the right track.

  26. anonymous
    • 4 years ago
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    the work force is equal to the PE.

  27. anonymous
    • 4 years ago
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    Work is equal to the same answer we got for number 3.

  28. anonymous
    • 4 years ago
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    because i was thinking w = f times d. and force = mass times acceleration i

  29. anonymous
    • 4 years ago
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    Now, from this value\[W = F \cdot y\]We can find the force from this expression, using the value from work obtained in number 5.

  30. anonymous
    • 4 years ago
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    so it would be 9800J = f * 80?

  31. anonymous
    • 4 years ago
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    Yes. I must leave now. I'll leave you with an outline of what to do: I must leave. I'll outline the solution to the remainder of the questions: 5. The amount of work comes from the definition of work I gave you earlier. \[W = \Delta U = \Delta KE + \Delta PE = \Delta PE\] 6. Comes from the other definition of work. \[W = F \cdot y\]We know the work from part 5, and the distance we move in y (100 - 20). We can solve for y. 7. Weight is equal to (mass)x(gravity) 8. Because we are not accelerating the object upwards in a true sense, we are just opposing gravity to lift the object. 9. \[P = {W \over t}\]

  32. anonymous
    • 4 years ago
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    thank you so much for your help ! i appreciate it !

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