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Homework Help: Work and kinetic friction coefficient

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A block slides down a frictionless hill, h=1.8m high. At the top of the hill the block has a speed of v=2.8m/s. At the bottom of the hill, the block travels over a patch of rough pavement and comes to a complete stop after s=5.0m distance due to friction.

    a) What is the coefficient of kinetic friction between the pavement and the block at the bottom of the hill?

    b) What is the magnitude of the averge acceleration of the block while traveling over the rough patch?

    2. Relevant equations
    Wf=ukmg x s
    Work energy theorem: W=0.5mv22-0.5mv12

    3. The attempt at a solution
    It seems that something is missing from this equation. Isn't mass a necessity here? Also, I'm not sure where to factor in the height of the hill into this equation.
  2. jcsd
  3. Mar 29, 2009 #2
    First try to calculate the block's speed at the bottom of the hill.

    Hint: the mass of the block is irrelevant - it will cancel out. Try it and see what happens.
    Last edited: Mar 29, 2009
  4. Mar 29, 2009 #3
    Conservation of energy

    The first step is to calculate the block's total energy at the top of the hill. The hill is frictionless, so the block's energy is conserved as it slides down the hill. That means you can calculate the block's speed when it reaches the bottom of the hill.
    Don't worry about that yet. Just start working on the problem and see what happens.
  5. Mar 29, 2009 #4
    Okay, I think I've gotten the kinetic friction coefficient by using the formula s=v2/uk2g which gave me 0.08.

    I am however a tad lost when it comes to determining average acceleration.
  6. Mar 29, 2009 #5
    I approached this problem differently than you - I found the average acceleration first.

    But anyway, you're using that formula incorrectly, because that's from your speed at the TOP of the hill.

    You need the speed at the bottom. Then you will be able to use your formula.
  7. Mar 29, 2009 #6
    then I'm lost ... because it seems like there are variables missing. If I'm to solve for average acceleration, don't I need an angle so I can determine the length of the hill?
  8. Mar 29, 2009 #7
    first the hill, then friction

    You can't solve this problem in one step. There are two parts to it: accelerating down the hill, and then slowing down at the bottom. You need to find the block's speed when it reaches the bottom of the hill before you can analyze what happens to it in the rough patch.
  9. Mar 29, 2009 #8
    Sorry, I think I confused you. Forget about the average acceleration for a minute.

    Let's first just worry about the speed at the bottom of the hill. You do this using conservation of energy calculations.

    Remember, total initial energy = total final energy.

    What kinds of energy does the block have at the top of the hill?

    What does it have at the bottom?
  10. Mar 29, 2009 #9
    At the top of the hill, the block has kinetic energy and at the bottom it has potential energy due to lack of movement.
  11. Mar 29, 2009 #10
    No, not quite.

    Whenever something is in motion (aka whenever it has a speed) it has kinetic energy.

    Whenever something has height, it has gravitational potential energy.

    At the top of the hill, does the block have motion? Yes, so it has kinetic energy. Does it have height? Yes, it is 1.8 m in the air. So, it has gravitational potential energy.

    At the bottom of the hill, the block still has movement! It doesn't just stop when it gets to the bottom - it continues moving for 5 seconds, right? So it has kinetic energy.

    But - the block has no height anymore! If it has zero height, it has zero potential energy.

    How can you put this information into a formula?
  12. Mar 29, 2009 #11
    k ... so it goes a little something like this then: K1 + Ugrav = K2.

    The mass cancels on both sides leaving (0.5vi2+gh)=0.5vf2 where the velocity at the bottom of the hill is 6.57m/s.
  13. Mar 29, 2009 #12
    Yes, that's right!

    (Be careful, though - what you have at the bottom of the hill is a speed, not a velocity).

    Now, you can use that formula you had before to find the coefficient of kinetic friction.

    You can find acceleration, too, because you know the block starts with a speed of 6.57 m/s, travels 5.0 m, and ends with a speed of 0 m/s.
  14. Mar 29, 2009 #13
    coefficient = 0.44

    average acc. = (6.57m/s-0m/s)/0.76s = 8.64m/s2
  15. Mar 29, 2009 #14
    0.44 is correct, but your acceleration isn't. Double check that time you found.
  16. Mar 29, 2009 #15
    am I using 6.57 and 0 as v1 and v2 respectively?
  17. Mar 29, 2009 #16
  18. Mar 30, 2009 #17
    Thanks for all your help!!!
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