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Help understanding answers: kinetic friction

  1. Oct 19, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    https://scontent-dft4-1.xx.fbcdn.net/v/t35.0-12/14796269_10210868496286479_1115393743_o.jpg?oh=0ccd692c60a1d0bab84d5fbaf7e6197b&oe=580A1894
    In the figure, a 6-kg box is sliding up an incline that makes an angle of theta=30 degree (with respect to the horizontal). As it slides up the incline, a kinetic frictional force fk acts on it. The acceleration of the block has magnitude 8 m/s2. Let the x+ direction point up the plane and the y+ direction be perpendicular to the plane.
    (A) draw free body diagram **I did this
    (B) What of the following statements correctly describes the x component ax of acceleration and the y component ay of acceleration as the block slides up the inclined plane? **answer: ax is negative, while ay is zero
    (C) What is the normal force acting on the box as it slides up the inclined plane? **answer: FN = 51.96 N
    (D) What is the kinetic frictional force acting on the box as it slides up the inclined plane? **answer: 18 N

    I do not need the answers. I only need help understanding why the answer is the answer for part (B) and help finding out how the answer was gotten for part (D).
     
    Last edited: Oct 19, 2016
  2. jcsd
  3. Oct 19, 2016 #2
    You forgot to upload the figure.
     
  4. Oct 19, 2016 #3
    I thought it was pretty straight-forward; I have attached it, though. Sorry.
     
  5. Oct 19, 2016 #4
    Part B: The problem defined the +x direction to be up the plane. And the problem states that the block is sliding "up the incline". So we know that its initial velocity is also in the +x direction. We don't really care about that for this problem though. When you sum the forces in the x direction, what direction is the resultant force (based on the sign convention specified in the problem)? In the positive direction or in the negative direction. The acceleration will be in the same direction as the resultant force.

    For Part D, it basically comes down to summing the forces in the x direction. Typically, when you sum all of the forces and you know the mass of the object, you can solve for the acceleration using one of those famous equations. In your case, one of the forces is an unknown, but you are given the mass of the object and its acceleration. So you can solve for the unknown force.
     
  6. Oct 19, 2016 #5
    Here's a similar situation. The +x direction is uphill and the -x direction is downhill in this case also. You are in your car and you are stopped with your car facing uphill. You step on the gas and start accelerating up the hill. Wouldn't your acceleration be positive since you are accelerating up the hill.

    Now let's say that when you hit 50 mph, you put your car in neutral while you are still going up the hill. What direction (and sign) is your acceleration now?
     
  7. Oct 19, 2016 #6
    Yes, it was straight forward. I thought the figure might contain some important information.
    Well, now tell me what is the net force on the object.
     
  8. Oct 19, 2016 #7
    Yes. You would then be rolling backwards, so, negative.
     
  9. Oct 19, 2016 #8
    The unknown force being Frictional, right? That's what I'm struggling with. I tried to do this:
    uk (mu sub k) = tan (theta) = tan (30 deg.) = 0.577350269

    fk (f sub k) = uk (mu sub k) * FN (F sub N) = (0.577) * (51.96 N) = 29.99911999

    The answer to the problem is 18 N, which is obviously not what I got.
     
  10. Oct 19, 2016 #9
    Can't I not find that, Net Force, until I find the fk (f sub k) or the kinetic friction.? Which is what I'm struggling to figure out..
     
  11. Oct 19, 2016 #10
    You wrote: "Yes. You would then be rolling backwards, so, negative."

    Immediately after you put your car in neutral while traveling 50 mph, you would still be rolling forward so your velocity would be positive, but your acceleration would be negative.

    I really messed up on trying to quote your post the first time. :/
     
  12. Oct 19, 2016 #11
    Calculate the net force in terms of mass and coefficient of friction.
     
  13. Oct 19, 2016 #12
    Ohhh, okay, yes. I see. Now, that is an example where the object starts and stops, but this example is where it just says that the object is pushed up...
     
  14. Oct 19, 2016 #13
    Where does it say that it is "pushed up"? I think that is an assumption that you made. It actually says it is "sliding up". "Pushed up" implies there is a force acting on it in the +x direction. But it doesn't say that.
     
  15. Oct 19, 2016 #14
    It's like the car that is still rolling uphill but it is in neutral. Obviously, some force had to have initially acted on the box to get it to be moving up the incline. But that force is no longer in effect in this problem.
     
  16. Oct 19, 2016 #15
    I'm just so confused on how a crate, with no legs/arms/feet/engine/force pushing it up, can be "sliding up" and incline.
     
  17. Oct 19, 2016 #16
    Okay, so it is just no longer in effect.... Okay...
     
  18. Oct 19, 2016 #17
    So in your problem, you basically have 2 forces acting in the x direction: the friction force and the x-component of the weight of the box. The x and y axes were conveniently chosen in this problem so that you don't really have to worry about the y direction. The acceleration (and velocity) has to be along the x axis - whatever direction they happen to be.
     
  19. Oct 19, 2016 #18
    Okay I get that. Now what?
     
  20. Oct 19, 2016 #19
    What equation do you know that relates force and acceleration?
     
  21. Oct 20, 2016 #20
    Force = mass * acceleration
     
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