1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Box sliding down plane - find work

  1. Nov 17, 2007 #1
    A 3.00kg box, initially at rest, slides 1.50m down a frictional plane inclined at 20degrees to the horizontal?



    a) The work done on the box was ?
    B) the final velocity of the box?




    a= 3kg*9.8m/s *sin(20)=10.05N
    10.05N *1.5m= 15.08N*m

    b= square root of 2* 10/3 * 1.5 = 3.16m/s
     
  2. jcsd
  3. Nov 17, 2007 #2
    In the question above, if the coefficent of friction is 0.250
    a) The work done on the box was?
    b)the final velocity of the box was ?

    Do I set it up 3*g*cos(20)= frictional
     
  4. Nov 17, 2007 #3

    Shooting Star

    User Avatar
    Homework Helper

    You have written "a= 3kg*9.8m/s *sin(20)=10.05N", which is not correct. This is not accn, but the component of mg along the plane.

    Total force along the plane should be mg*sin(20)-(mu)mg*cos(20). Now you can find W.
     
  5. Nov 17, 2007 #4
    what is mu
    and why can you tell all the equation you used.
     
  6. Nov 17, 2007 #5

    Shooting Star

    User Avatar
    Homework Helper

    mu, pronounced mew, is the greek letter we use to denote the co-eff of friction. It's not m into u. We'll use 'k' for it hence.

    Frictional force F = kN, where N is the normal reaction at the point of contact.

    Normal reaction is mgcos(20 deg) in this case. Total force P acting on the body along the plane is component of weight along the palne minus frictional force. So,
    P= mg*sin 20 – kmg*sin 20.

    W = F*d.

    I am sure you know the formula for final velo and accn. Find the accn and apply it.
     
  7. Nov 17, 2007 #6
    Ok in the first question , you said used the equation
    mg*sin(20)-(mu)mg*cos(20). And mu=frictional force. However in the first equation we are to asumed there is no frictional force because it is no stated. So this equation is used for the second question where frictional force is given?
     
  8. Nov 17, 2007 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ?? Why do you say there is no frictional force? What you wrote was, "A 3.00kg box, initially at rest, slides 1.50m down a frictional plane inclined at 20degrees to the horizontal?" Did you mean "frictionless"?
     
  9. Nov 17, 2007 #8
    for the second part, you should look carefully at what the acceleration should be.. is it really 9.8 or 10(as u wrote it) or something else, maybe a component of gravity.
     
  10. Nov 18, 2007 #9

    Shooting Star

    User Avatar
    Homework Helper

    You have not given the co-eff of friction. Is it a frictionless plane? In that case, k=0. The component of the force along the plane you know. Divide my m to get the accn a. You also know the initial velo.
     
  11. Nov 18, 2007 #10
    Okay, I am sorry for the confusion. This is a two part question the first part is frictionless, and it states..

    A 3.00kg box, initially at rest, slides 1.50m down a frictionless plane inclined at 20degrees to the horizontal?

    a) The work done on the box was ?
    B) the final velocity of the box?

    and I set it up like this

    part a=
    3kg*9.8m/s *sin(20)=10.05N
    10.05N *1.5m= 15.08N*m

    b= square root of 2* 10/3 * 1.5 = 3.16m/s
     
  12. Nov 18, 2007 #11
    The second part has friction and the question is

    In the question above, if the coefficent of friction is 0.250
    a) The work done on the box was?
    b)the final velocity of the box was ?

    And I set it up like

    part a =
    3kg*9.87*sin(20) = 10.05
    3kg*9.87*cos(20)= 27.62

    Un= (.250)*(27.62)= 6.9
    10.05-6.9-3.15

    N= 3.15/(.250)= 12.6N


    part b

    v^2= vo^2+ 2aX

    v= square root of 12.6/3 *1.5= 2.51


    is this right
     
  13. Nov 18, 2007 #12

    Shooting Star

    User Avatar
    Homework Helper


    This one seems all right. Why 10 instead of 10.05? But let it go.
     
  14. Nov 18, 2007 #13

    Shooting Star

    User Avatar
    Homework Helper

    What is N? You got the force as 3.15.
     
  15. Nov 18, 2007 #14
    I was following a homework example, I must have wrote down N as a mistake.
     
  16. Nov 18, 2007 #15

    Shooting Star

    User Avatar
    Homework Helper

    Well, correct it. The force is 3.15. Then proceed as in the first part.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Box sliding down plane - find work
Loading...