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Inclined plane forces problem

  1. Jul 10, 2013 #1
    A 2.0kg wood block is launched up a wooden ramp that is inclined at a 32 degree angle
    Initial speed of the block is 8 m/s and the coefficient of kinetic friction is 0.2.

    a) what height does the block reach above its starting point?

    b) What speed does it have when it slides back down to its starting point?

    We have to do this without the Work-Energy Thorem
    Only using Forces

    For part a, I summed the forces in both the x and y direction and solved for the acceleration. I then plugged the acceleration into the V2= Vi2 + 2 a y equation, assuming final velocity was 0, but my answer was still wrong.

    Sum of Fx forces: -fk - mgsintheta = ma

    Sum of Fy forces: N - mgcostheta = 0

    What am I doing wrong? Please answer. Am I supposed to find t first?
     
  2. jcsd
  3. Jul 10, 2013 #2

    TSny

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    Hello, Chemistrystrng. Welcome to PF!

    Your work looks fine to me. You do not need to find t. We would need to see more detail of your calculation to know if there is a mistake somewhere in the calculation. The question is a little ambiguous: does "height" mean distance along the incline or vertical height?
     
  4. Jul 10, 2013 #3
    Hi :)
    It's the vertical height. I forgot to specify. So what vertical height does the block reach above its starting point? for part a

    So I drew the FBD, and summed the forces in both x and y direction as shown above. Once I did that I got -ukN - mgsintheta = ma, so Normal fore is mgcostheta.

    Thus, making it -ukmgcostheta - mgsintheta = ma. So I divided by mass and solved for acceleration which is -6.86 m/s2. But when I plug it into the kinematics equation, both the one with t and the V2 = Vi2 + 2ay one, I get the wrong answer. So I don't know what it is I'm doing incorrectly.
     
  5. Jul 10, 2013 #4

    TSny

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    I agree with the value you got for the acceleration. What did yo get for the vertical height?
     
  6. Jul 10, 2013 #5
    I got 7.71 m..but I don't know if this is correct or not. I assumed final velocity was zero and I used 8sin32 to calculate the initial velocity in the direction of the object. But what I don't understand is that it's 0-8sin32^2, making it positive, and since acceleration is negative, the answer is a negative 7.71m..so I don't know what I'm doing wrong...
     
  7. Jul 10, 2013 #6
    Wait, no i did not get that. I got -1.3 m...and I know that's not right. But I used V2 = Vi2 + 2ay...
     
  8. Jul 10, 2013 #7

    TSny

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    I'm not sure what you mean by "velocity in the direction of the object". But, you can assume that the 8 m/s is the initial speed of the block along the incline.

    What in particular is positive? The minus sign is not squared here.
     
  9. Jul 10, 2013 #8
    Oh, so you don't need to use 8sin32? It's just 8 m/s? If so, then the height is 4.7, if the negative sign is not squared. But that is also incorrect...

    By making it positive, I meant the V2 - Vi2
     
  10. Jul 10, 2013 #9

    TSny

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    Does the 4.7 m that you got represent the distance along the incline or the vertical height?
     
  11. Jul 10, 2013 #10
    The distance along the incline...so you would do 4.7sin32 to get the vertical height..giving 2.5 m.

    Thank you so much! I thought the 4.7 was the vertical height.

    I have trouble with physics problems...I understand the concepts, but I cant do the actual problems without outside help. It's difficult for me. I do thank you for this. I guess I'll have to practice some more
     
  12. Jul 10, 2013 #11
    And for part b, we just use the V2 equation since we have the incline height? Or am I completely wrong again?
     
  13. Jul 10, 2013 #12

    TSny

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    In your original post, you summed the forces along your x-axis to get ax. Since you chose the x-axis along the incline, ax is the acceleration along the incline. Then when you use vx2 - vox2 = 2axx to solve for x, you are getting the distance along the incline.

    I also get 2.5 m.
     
  14. Jul 10, 2013 #13

    TSny

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    Well, yes, but you better be careful with the direction of the friction force coming back down.
     
  15. Jul 10, 2013 #14
    Okay, so as it's going down, would the forces in the x direction be -mgsintheta + fk? and then I substitute the fk for umgcostheta and solve for acceleration, correct? same as part a except we have the 2.5 m height this time, right?
     
  16. Jul 10, 2013 #15

    TSny

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    Yes, that sounds good.
     
  17. Jul 10, 2013 #16
    I got the answer as 5.7 m/s and it is correct. Using the 4.7 incline height. But, is the acceleration positive or negative as it goes down? Because isn't the initial velocity 0? And if so, would the signs of those two forces mentioned above be different?
     
  18. Jul 10, 2013 #17

    TSny

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    You chose the positive x-direction to be up along the incline. When the block comes back down the incline, it will increase its speed in the negative x-direction. (That is, the velocity becomes more and more negative.) That should tell you what to expect for the sign of ax.

    [EDIT: Yes, the two forces along the incline are in opposite directions when the block is sliding back down. So, they should have opposite signs.]
     
  19. Jul 10, 2013 #18
    Oh, okay so the velocity is negative, and so is the acceleration since it's speeding up. But when using the V2 equation, it comes out to be V2 = sqrt (-33.276) since initial velocity is zero (I think). Do I make the acceleration positive since it's asking for speed and not velocity?

    I'm slow =(
     
  20. Jul 10, 2013 #19

    TSny

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    The square root of a negative number does show that something is wrong. Did you take into account the sign of the displacement coming back down? (Note: in the equation vx2 = vox2 + 2axx, x represents the displacement from the starting point. It would probably be better to write the equation as vx2 = vox2 + 2axΔx where Δx is the displacment from the point where the velocity is vox to the point where it is vx. When the block is coming back down, is the displacement positive or negative?

    Not at all. These signs are very tricky to deal with.
     
  21. Jul 11, 2013 #20
    Oh, yes. It is coming down the slope, so it's final - initial, so 0-4.7, so it's -4.7, which would take care of the - acceleration. Giving the answer to be a positive 5.7 m/s.

    Thank you so much. That helped a lot.
     
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