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Work and Energy Related Problem

  1. Nov 23, 2003 #1
    Okay, I have this problem that I've been having trouble with. It seems like a simple problem, but I keep getting the wrong answer and I don't know where I'm making the error. Maybe one of you can help me target the problem. =) The problem is as follows:
    Well, firstly I converted the 50.0 hp into watts by multiplying it by 746 (because one hp equals 746 watts). I got 37300 watts. Then using the formula [tex]P=\frac{W}{\Delta t}}[/tex], I plug in 6.40 x 105 J as P(ower) and 6.40 x 105 as W(ork). When I calculated it, I get .058 seconds for t. However, the correct answer is 17.2 seconds. Can anybody tell me where I'm going wrong and waht I need to do?
     
  2. jcsd
  3. Nov 23, 2003 #2

    Doc Al

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    Staff: Mentor

    You must be plugging things in wrong. Try again (you have everything right):

    ΔT = W/P
     
  4. Nov 23, 2003 #3
    Ohhh... haha. Whoops, I was just making a calculation error. Thanks! Okay I get it. I have another question though. In the following problem:
    Okay, for this problem, I'm not quite sure how to approach it. Initally, I thought I had to simply use the formula [tex]P=\frac{Fd}{\Delta t}}[/tex] to solve it by pluggin in 4.0 x 103 N as F and 3.00 m as d, and jsut put in one second for t. However, when I calculated it, it didn't come out to the correct solution so I'm assuming that this approach was not right. How then, would I solve this problem?
     
  5. Nov 23, 2003 #4

    AD

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    You know that P = F(Δd/Δt)

    Δd/Δt is equal to velocity, so P can also be expressed

    P = Fv

    The force the lift motor has to overcome when the lift is fully loaded is

    F = (1000 + 800)g + 4000

    So

    P = (1800g + 4000)v

    You didn't put the value of v in your question, but you mentioned that the lift moves 3m in one second, so

    P = 3(1800g + 4000)
     
  6. Nov 23, 2003 #5

    Doc Al

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    Staff: Mentor

    You forgot that the motor has to work against gravity, not just the friction.
     
  7. Nov 23, 2003 #6
    What do you mean when you refer to 'g'?
     
    Last edited: Nov 23, 2003
  8. Nov 23, 2003 #7

    AD

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    The constant g is the acceleration due to gravity.
     
  9. Nov 23, 2003 #8
    So then would I have to multiply 1800 by 9.81 m/s2?
     
  10. Nov 24, 2003 #9

    AD

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    Yes.
     
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