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Pretty easy, just want someone to verify I have the right answer

  1. Sep 25, 2013 #1
    Pretty easy, just want someone to verify I have the right answer:)

    1. The problem statement, all variables and given/known data

    A worker pushes downward and to the left on the box at an angle of 35°.

    The box weighs 135 kg and moves horizontally across the floor to the left at a constant velocity.



    2. Relevant equations
    15. What is the value of the friction force?
    a.1164 N
    b.1663 N
    c.2489 N
    d.1324 N


    16. What is the value of the acceleration of the box?
    a. 6.7 m/s2
    b. 0 m/s2
    c. 18.4 m/s2
    d. 12.3 m/s2





    3. The attempt at a solution
    I worked out the numbers and arrived at 1324 for the first one and 6.7 for the second, but I would love for someone to verify this before I submit it :)
     
  2. jcsd
  3. Sep 25, 2013 #2
    I do not think there is enough information to answer 15. For 16, what does "constant velocity" imply?
     
  4. Sep 25, 2013 #3
    What?The angle which you gave is from horizontal or vertical?..And if the block moves horizontally across the floor to the left at a constant velocity,then how can it have acceleration?(Direction of velocity is also not changing so no production of acceleration due to change in direction of velocity)..Rephrase the question and check your answers?
     
  5. Sep 25, 2013 #4
    You are right

    You are right.The magnitude of applied force is missing.
    Just asking that if the force applied is given then shouldn't the x-component of the applied force be equal to the frictional force if the block is not accelerated?
     
  6. Sep 25, 2013 #5
    μs=0.4
    μk=o.2. My bad for leaving that out. Does that help?
     
  7. Sep 25, 2013 #6
    It should, but we are not given the magnitude of the applied force, nor the coefficient of friction. Knowing either one would make the problem determinate.
     
  8. Sep 25, 2013 #7
    What is the angle relative to?
     
  9. Sep 25, 2013 #8
    It may solve..working on it.but tell the angle
     
    Last edited: Sep 25, 2013
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