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Calculate the average force acting on the block

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A block of mass 2 kg starts from rest and begins to move at time t=0 along a straight line on a smooth horizontal surface. The displacement of the block, s, depends on time, t, given by s=3+2t+t^3 . t is expressed in second and s in meter. Calculate the average force acting on the block during the first 3 seconds of its motion. Express your answer in Newton.

    2. Relevant equations
    F=ma


    3. The attempt at a solution

    To find avg force , I'll need to find avg acceleration and then multiply it with mass of block.
    But the problem is I don't know how to calculate avg acceleration from displacement as a function of time . Please Help.
     
    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2
    What is the definition of average acceleration?
     
  4. Jun 25, 2013 #3
    Avg acceleration = v-u/t
     
  5. Jun 25, 2013 #4

    collinsmark

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    Is this a calculus based physics class?

    What's the relationship between instantaneous velocity and instantaneous displacement?
     
  6. Jun 25, 2013 #5

    haruspex

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    (v-u)/t, yes. You know u; how will you calculate v? (See collinsmark's hint.)
     
  7. Jun 26, 2013 #6
    Instantaneous velocity is derivative of displacement, so v= 2+3t^2
    so at the third second v= 2+3*9=29 .
    But there are 2 problems , first when I substitute 0 for t I get v=2 But they told us that the object starts from rest ! Also I'm probably not supposed to use calculus . ( Its OK if I have to , but if there is a way without calculus then it is preferred .)
     
  8. Jun 26, 2013 #7

    collinsmark

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    That looks right.

    That is a good point.

    You might want to check to make sure the equation was typed in/written down correctly. Either something is wrong with the s = 3 + 2t + t3, or something is wrong with the statement that the object starts from rest (with the object beginning to move at time t = 0). One way or the other, something is wrong with the problem statement.

    Calculus or no calculus, it doesn't change the fact that the object has a velocity of 2 m/s at time t = 0, if we are to believe the
    s = 3 + 2t + t3 equation. Using a non-calculus method won't change that fact. Calculus makes doing physics problems easier, but it doesn't change any results.

    [Edit: One way out of this (sort of) is if the object was hit with an impulse at time t = 0, such that there was infinite force applied over an infinitesimal interval of time, and the area of the impulse is finite (kind of like a baseball being hit with a baseball bat). With this, you can still calculate the average force from time t = 0 to time t = 3, and the result will be finite. But you will get different answers depending on whether you start at 0+ vs. 0-. So something still looks fishy to me in the problem statement.]
     
    Last edited: Jun 26, 2013
  9. Jun 27, 2013 #8
    Maybe the velocity at t=0 is really 2 because when I reread the question , I realised that they say the object starts from rest BUT starts to move at t=0. So lets suppose that v=2 at t=0 .
    Then acceleration (avg) =29-2/3=9.
    So avg force = 18 , right ? But the answer sheet says that the correct answer is 16.
    Am I wrong or is the answer sheet wrong? Please tell quickly .
     
  10. Jun 27, 2013 #9
    Sounds way too cryptic for something that ought to be easy. Most likely, a mistake in the formulation.
     
  11. Jun 27, 2013 #10

    haruspex

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    I tried to figure out what the mistake is in the question. To resolve the initial speed paradox, need to get rid of the linear term, 2t, in the s equation. Maybe 2t^2? That gives 26 as the answer. Two typos in one question?
     
  12. Jun 28, 2013 #11
    Forget about that , lets suppose the object doesn't start from rest (but with v=2) .
    Now what answer do you get ? is it 18? Please conform the answer asap.
     
  13. Jun 28, 2013 #12

    collinsmark

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    Don't forget your units, but yes, that looks right.

    On the other hand, if the block does experience an impulse at t = 0, and this is to be taken into account (i.e., the block starts from rest), you'll get a slightly different answer (but it's still neither 18 nor 16 N). Modeling such an impulse isn't too far fetched. All it means is that if you plot velocity vs. time there will be a discontinuity at t = 0: The curve will jump from 0 m/s to 2 m/s at that point. The area under the curve at time t gives you the block's position. The slope of the curve at time t is the block's acceleration. (The impulse comes into play when trying to determine the slope at time t = 0.)

    But as others have pointed out, it sort of looks like the author of the problem got confused and put the wrong equation into this problem statement.
     
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