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Work done to crate on angle

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A 128 N carton is pulled up a frictionless baggage ramp inclined at 30.0 degrees above the horizontal by a rope exerting a 72.0 N pull parallel to the ramp's surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by the rope, gravity, and the normal force of the ramp.

    2. Relevant equations

    W= force*cos(theta)* displacement

    3. The attempt at a solution

    Work done on carton by ROPE W=72cos(30)*5.2 Answer comes out to 324 J, which is wrong.

    Work done on carton by GRAVITY W=128cos(30)*5.2 Answer comes out to 576 J, which is wrong...

    What am I getting incorrect about this????
  2. jcsd
  3. Oct 26, 2008 #2
    Well, i figured out the work done on the carton by the rope because the angle of the rope is NOT 30 degrees, it is parallel, thus 0degrees. Still can't figure out the gravity one though..
  4. Oct 26, 2008 #3
    Did you start with a sketch showing all of the forces acting on the object, along with the ramp and the angle theta?

    Where did you get this equation from? What do the variables in it represent? Is the equation relevant here?
  5. Oct 26, 2008 #4
    NVM, figured them all out!!
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