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Work of pushing box up ramp

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose you lift a 23kg box by a height of 1.0m. How much work do you do in lifting the box?
    230 J

    Instead of lifting the box straight up, suppose you push it up a 1.0 -high ramp that makes a 30 degree angle with the horizontal. Being clever, you choose a ramp with no friction. How much force is required to push the box straight up the slope at a constant speed?
    113N

    How long is the ramp?
    2m

    Use your force and distance results to calculate the work you do in pushing the box up the ramp.
    This is where I'm having trouble!

    2. Relevant equations

    W=Fdcosθ

    3. The attempt at a solution

    I plugged what I already got into the above equation, and got the answer of 195J. However, it's telling me the answer is 230J? But how? Isn't that why they had me find F and d?

    Thank you for any help!
     
  2. jcsd
  3. Dec 6, 2012 #2
    All your work is correct apart from the value for cosθ.
    What is the angle between the force acting up the ramp and the displacement up the ramp?
     
  4. Dec 6, 2012 #3
    I'm sorry, I really don't know. What angle between the force and displacement? I thought that was the 30 degrees.
     
  5. Dec 6, 2012 #4

    haruspex

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    In that formula, θ is the angle between the force and the distance through which it acts. You must resist the temptation to think "this formula needs an angle, I have an angle, so I'll put in the formula".
    The force in this case is in which direction?
    The distance the box moves is in which direction?
    What is the angle between those two directions? (Hint: it isn't much.)
     
  6. Dec 6, 2012 #5
    When you push the box along the ramp then the force you apply and the displacement of the box are both along the ramp. So the angle between them is zero, not 30°
     
  7. Dec 6, 2012 #6
    Ohh, ok, I got it now, thanks! :) I wasn't even thinking about that. Like haruspex said, I thought it was 30 degrees because I thought that's what it was referring to.

    @ haruspex:
    The force is to the right.
    And the displacement of the box is to the right.
    So, the angle between them must be zero?

    So, in other words, the angle is zero regardless that it's on a ramp, because the force applied to the box wasn't at an angle to the box? I hope I said that right.
     
  8. Dec 6, 2012 #7

    haruspex

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    Need to be more precise than that. The direction of the force (as you have calculated it) is directly up the slope of the ramp. The direction of movement is also in exactly that direction. So, yes, the angle between is zero.
    Note that the pusher could have pushed in a different direction, horizontally, say. You would now calculate a larger force (because it would increase the normal force from the ramp). If there were friction then this would mean it would take more energy to move the box up the ramp. But since there is no friction, the component of the force up the ramp is unchanged:
    A)
    force directly up ramp = 113N
    distance moved = 2m
    angle between = 0
    work done = 113*2 cos(0) = 226Nm
    B)
    horizontal force = 113N / cos(30)
    distance moved = 2m
    angle between = 30 degrees
    work done = (113/cos(30))*2* cos(30) = 226Nm
     
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