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Posted by on Wednesday, November 16, 2011 at 5:58pm.

A box and its contents have a total mass M = 9 kg. A string passes through a hole in the box, and you pull on the string with a constant force F = 77 N (this is in outer space; there are no other forces acting).





(a) Initially the speed of the box was vi = 5 m/s. After the box had moved a distance w = 2.1 m, your hand had moved an additional distance d = 0.4 (a total distance of w + d), because additional string of length d came out of the box (see figure). What is now the speed vf of the box?

  • physics - , Saturday, November 17, 2012 at 8:17pm

    you do
    Fd=.5m*vfinal^2-.5m*vinitial^2
    solve for vfinal and that is the correct answer.

  • physics - , Wednesday, November 13, 2013 at 6:11pm

    sjdsj

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