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Homework Help: Determine the work done

  1. Mar 31, 2017 #1
    1. The problem statement, all variables and given/known data
    An object is subject to a nonconstant force f = (6x^3 -2x) ihat

    such that the force is in newtons when x is in meters. Determine the work done on the object as a result of this force as the object moves from x = 0 to x = 100 m

    2. Relevant equations
    w = f * d

    3. The attempt at a solution
    again not sure what I'm doing wrong here. Plugging in 100 m into f you get

    5999800 m, and the total distance covered was 100 m, so multiplying the two you get 5.999 x 10^8 joules but book gives me 1.5 x 10^8 joules

    I do not see how the books answer agrees with the equation w = f * d.

    alternately I could use [itex] W = \Delta K + E_t [/itex] but I don't think using this equation is right in this case...
  2. jcsd
  3. Mar 31, 2017 #2


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    Staff: Mentor

    I'm pretty sure you need to use an integral for this problem. When the force changes as a function of distance, you need to integrate the force f(x) * dx over the range of distance x. Can you show us that work?

    If you haven't use LaTeX before to show integral work, you can find a tutorial under INFO at the top of the page (click Help/How-To).
  4. Mar 31, 2017 #3
    I see. So if acceleration is not constant, then we have to use integral method? and if the force IS constant, we can use equation W = F*d?
  5. Mar 31, 2017 #4
    This involves basic calculus.. you need to find the Work equations first which is

    6/4 × X^4 + X^2

    now insert X = 100 and X = 0

    1.5 10^8 +10^4 - (0)^4 -0^2

    Wait the answer is different from your book, why?._.
    (Btw i can't give you the proper equations since im on mobile, its too messy)
  6. Mar 31, 2017 #5


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    Staff: Mentor

    I'm not sure this is helpful. Let's see how the OP's work looks first...
  7. Mar 31, 2017 #6


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    Staff: Mentor

    You haven't been given any information about the acceleration. You are given f(x) and asked to find the work done over a range of x. That means you need to do the integral to figure out the total work over that distance.
  8. Mar 31, 2017 #7
    Hmm so I integrate function 6x^3 -2x

    I get (6/4)x^4 - x^2
    (6/4)(100)^4 - (100)^2 = 149990000

    so that is my force. My distance is 100, so I multiply 149990000 by 100, and get 1.5 x 10^10

    Still not my books answer :(. Why am I off by 10^10 - 10^8 :(???
  9. Mar 31, 2017 #8
    Well its saying that the force is not constant, so I assumed acceleration to be non constant as well... does that seem right?
  10. Mar 31, 2017 #9
    No no, you got it right. Wow the answer was much closer to the book than i thoguht.

    So,149990000 Joule IS your Work. Which rounds up to 1.5 × 10^8. The integral of Force IS work. Your equation has been converted to work equation
  11. Mar 31, 2017 #10
    Ahh okay finally I understand now.

    Using the fact that integral of force is work, I didn't have to multiply by 100 at all.

    Okay thank you guys. But also if you can answer my question:

    "Well its saying that the force is not constant, so I assumed acceleration to be non constant as well... does that seem right?"

    I appreciate everything!
  12. Mar 31, 2017 #11
    Well yes.. if your Force changed, either the accel. or the mass changed. But you dont gain/lose mass along the way :p so yeah acceleration is the one that changes
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