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physics

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An archer puts a 0.26-kg arrow to the bow-string. An average force of 200 N is exerted to draw the string back 1.4 m.
(a) Assuming no frictional loss, with what speed does the arrow leave the bow?
m/s
(b) If the arrow is shot straight up, how high does it rise?
m

  • physics - ,

    1.
    W =F•x =200•1.4 =280 J,
    W =KE = m•v²/2,
    v =sqrt(2•W/m) = sqrt(2•280/0.26) =46.4 m/s.
    KE =PE = m•g•h
    h = KE/m•g = 280/0.26•9.8 ≈110 m.
    2.
    Sometimes I’ve found the answers of this problem based on the following solution
    PE =kx²/2 = KE =mv²/2.
    This solution gives v =sqrt(k•x²/m) = sqrt(F•x²/x•m) = sqrt(F•x/m) = 32.8 m/s.
    And the height from m•g•h =m•v²/2 is
    h =v²/2•g = 54.95 m.
    But I believe that this solution is incorrect because we have given the magnitude of the average (!) force
    So the first solution is quite the thing.

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