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Acceleration of a Rock Problem

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    A 0.0200-kg rock is shot directly upward with an average force of 98.0 n. The rock accelerates through a distance of 0.150 m.

    2. Relevant equations

    a.) Determine the acceleration while in the slingshot.
    b.) What is the velocity with which it leaves the slingshot?
    c.) What force acts on the rock after it leaves the slingshot and what is its value?
    d.) What is the value of the acceleration/deceleration after it leaves the slingshot?
    e.) How far upward does it travel before it begins to fall?
    f.) How long is the rock in the air upon returning to the level of the slingshot?


    3. The attempt at a solution

    I've tried v= v0+ at for part a; however, my Physics teacher said I was so wrong. Could you help with with this problem? I've tried everything, and I can get the rest of the worksheet. It's just this problem that makes my head spin around like Linda Blair...

    Thanks so much and have a great day!!!! :)

    Andrew
     
  2. jcsd
  3. Nov 10, 2011 #2

    gneill

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    Staff: Mentor

    Hi Andrewy77. Welcome to Physics Forums.

    What forces are acting on the mass? What's the total force? What's the formula for acceleration given total force and mass?
     
  4. Nov 10, 2011 #3
    Thanks for the greeting!

    The problem doesn't say; therefore, I think the only force acting on it is gravity, which is 9.8 m/s.

    Well, given that m= f/a

    a= f/m. That is all the information he gave us.
     
  5. Nov 10, 2011 #4

    PeterO

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    Homework Helper

    Newton's Laws; specifically the second.
    F = ma is what you are after - you have the force, you have the mass, you can find the [average] acceleration.
     
  6. Nov 10, 2011 #5

    gneill

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    Staff: Mentor

    Did you draw a Free Body Diagram for the rock while it's in the slingshot?
     
  7. Nov 10, 2011 #6
    Yes; the diagram helped a little.

    For part a), I used f=ma

    Solving for a gets you a=f/m

    With the given data, I got 4900 m/s for a. That sounds a little quirky..
     
  8. Nov 10, 2011 #7

    gneill

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    Staff: Mentor

    Looks like a plausible acceleration for such a small mass. However, I have one question that should be decided before you proceed: Is the given 98.0N force the net force on the stone including the effects of gravity (weight), or is it the average force applied by the slingshot alone? If its the latter case, then the actual acceleration will be reduced by g.
     
  9. Nov 10, 2011 #8
    Well, the problem states the rock was shot with 98.0 n of force, so I'm guessing it's the force applied to shoot the rock.
     
  10. Nov 10, 2011 #9

    gneill

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    Staff: Mentor

    If that is the case then subtract g from the acceleration you calculated above:
    [tex] a = \frac{f_{net}}{m} = \frac{F - m g}{m} = \frac{F}{m} - g[/tex]
     
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