Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work & energy; please help

  1. Mar 8, 2007 #1
    work & energy; please help!!!!!

    i have this one problem in my hw tht i cannot solve:

    A person pushes a 16kg cart at a constant velocity for a distance of 22.0m. She pushes in a direction 29 degrees below the horizontal. A 48 N frictional force opposes the motion of the cart. a) what is the magnitude of the force that the shopper exerts? b) determine the work done by the pushing force.

    please guide me or show me the steps


  2. jcsd
  3. Mar 8, 2007 #2
    It's always better to make an attempt at the solution. Heard anything of decomposition of forces? Since the velocity is constant, the horizontal (tangential) component of the person's force is equal in magnitude and opposite in direction to the frictional force.

    ad a) Make a drawing and then use trigonometry to get the total force the shopper exerts.

    ad b) Hope you know the definition of work, force * distance. Don't forget what force to calculate with. Is it the total force or just the horizontal (tangential) component?
  4. Mar 8, 2007 #3
    i have already made the drawing

    however i am not sure how to seperate it into its components

    also how do you mathematically express the horizontal component to the frictional force?

    and for part b it is just the horizontal component

  5. Mar 8, 2007 #4
    Well, it should be apparent from the drawing, so post it here and let us see what you've made so far.

    And yep, it's just the horizontal component (the component in the direction of the motion) that's used when calculating work.
  6. Mar 8, 2007 #5
    i kno tht the horizontal component is fcos29

    however i dont seem to be able to get to the force

    im completely lost

  7. Mar 8, 2007 #6
    Vertical and horizontal components of the total force form a rectangle. The total force is its diagonal - the angle between the horizontal force and the total force is 29 degrees. Everything else stems from this drawing.

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook