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How to find the cosine between the directions

  1. Nov 28, 2004 #1
    I really need some help on this one:

    You push a box up a ramp using a horizontal 100-N force F. For each 5m of distance along the ramp the box gains 3m of height. Find the work done by F for each 5m it move along the ramp (a) by directly computing the dot product from the components of f and the displacement s, (b) by multiplying the product of magnitudes of f and s with the cosine of the angle between their directions (c) by finding the component of the displacement in the direction of the force and mulitplying it by the magnitude of the force.


    I am pretty sure I know how to find the first part which is just
    W=F*(5m i + 3m J) which would be 100N*5m + 100N*3m but I have no clue on the other two parts. For part b I am not sure how to find the cosine between the directions, I know the magnitude of F and is the magnitude of s just 5m + 3m? Any hints are greatly appreciate as I do not even know where to start. Thankyou
     
  2. jcsd
  3. Nov 28, 2004 #2
    If I understand the question correctly then the work done in part (a) is given by [itex]W=\vec{F} \cdot \vec{s} = (100N\hat{\imath} + 0N\hat{\jmath}) \cdot (4m \hat{\imath} + 3m \hat{\jmath}) = 400J[/itex]. (b) [itex]W = Fs\cos\theta = (100 N)(5 m)\cos\theta[/itex]. I suppose they gave you an inclination angle [itex]\theta[/itex] of the ramp that you can substitute in. Otherwise you can use the value of [itex]F[/itex] calculated in part (a) to find [itex]\cos\theta[/itex]. (s) [itex]W = F_xs_x = (100N)(4m) = 400J[/itex]
     
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