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I need some help!

  1. Nov 19, 2005 #1
    I am not sure how I am supposed to do these problems and I am going to be tested on them soon can someone please help me.

    1) An 83 g arrow is fired from a bow whose string exerts an average force of 95 N on the arrow over a distance of 85 cm. What is the speed of the arrow as it leaves the bow? Do I use KE=.5m(v^2)?. If so how do I use that since I dont know the velocity?

    2)An automobile is traveling along a highway at 94 km/h. If it travels instead at 103 km/h, what is the percent increase in the automobile's kinetic energy? I have used KE=.5m(v^2) but I dont understand how I get the percentage!

    Please I need some major help!:cry:
     
  2. jcsd
  3. Nov 19, 2005 #2

    Astronuc

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    Staff: Mentor

    Well, assuming that the arrow starts at rest and then is accelerated by the force of the bowstring, the change in energy is just the integral of force over the distance over which the force is applied. However, since this an average force, one can simply use the product of force * distance, then use the relationship of kinetic energy to velocity and solve for the velocity.


    #2. Calculate the KE for each velocity.

    Then calculate the difference in the two KE's, i.e. [itex]\Delta[/itex]KE.

    The % difference is just [itex]\Delta[/itex]KE/KE1*100.

    Example, something is traveling at 100 mph when it accelerates to 110 mph.
    110 mph represents a 10% increase of 100 mph from (110 - 100)/100 * 100.

    100% = 1.
     
  4. Nov 19, 2005 #3
    I understand #2 but #1 is still very confusing to me and I dont know how to do it.
     
  5. Nov 19, 2005 #4

    Astronuc

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    Staff: Mentor

    Are you comfortable about the equivalence of Work and Energy.

    Remember Newton's Second Law - F = ma.

    And remember the kinematic equation which relates acceleration, a, to some displacement, [itex]\Delta[/itex]x,

    2a [itex]\Delta[/itex]x = vf2 - vf2.

    Multiplying this equation by m and dividing by 2 gives

    ma [itex]\Delta[/itex]x = 1/2 mvf2 - 1/2 mvf2, and applying Newton's law,

    F[itex]\Delta[/itex]x = KEf - KEi = [itex]\Delta[/itex]KE.

    This of course assumes that F is constant or an average value applied over some distance or displacement, [itex]\Delta[/itex]x.
     
    Last edited: Nov 19, 2005
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