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Force of 5N applied to 5kg block on table w/ μk = .1

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data

    A force of 5N is being applied to a 5kg block is on a table. The μk = .1.
    What is the velocity of the block?


    2. Relevant equations

    (μk)(Fn) = Ff

    3. The attempt at a solution

    (.1)(50N) = 5N of friction.

    The net force is 5N - 5N = 0N


    From here, I believe it is impossible to find the velocity.
     
  2. jcsd
  3. Oct 2, 2015 #2

    gneill

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    Staff: Mentor

    Why's that?
     
  4. Oct 2, 2015 #3

    Because net force = 0, thus F/m = 0N/5kg = 0 acceleration.
    With 0 acceleration, velocity is constant. In order to solve for velocity, you need to know the distance and time over which the block traveled.

    The only information I have available is the applied force, friction force, net force, and coefficient of kinetic friction. I don't know how find velocity from this information.

    I want say it's completely possible, but velocity is v = d/t. How can I turn the available information into distance and velocity values?
     
  5. Oct 2, 2015 #4
    let the distance be x
    and apply work-energy theorem
     
  6. Oct 2, 2015 #5

    gneill

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    You say the velocity is constant because the net force acting is zero. That's true! So you can state that the velocity is equal to the initial velocity, whatever that might be. Now, is there any hint in the problem statement that might lead you to a value for the initial velocity? Note that it might be a subtle hint...
     
  7. Oct 2, 2015 #6

    If I let distance be x, will I not have two unknowns in one equation? Also should Force be Net Force or Applied Force. If net force is used, then since net force is zero, it would make it impossible to determine velocity, right? But even using applied force, I get different velocity values when I plug in random distance values as I believe you suggested. Not exactly sure what to make of it:

    [itex] F*d = \frac{1}{2}mv^{2} [/itex]
    [itex] 5N(x) = \frac{1}{2}(5kg)v^{2} [/itex] now I would have to solve for both x and v?

    Even if I put in any number for distance, say 10 meters. v = [itex] sqrt{20}[/itex]
    But, if I let distance be 100 meters, v = [itex] sqrt{200}[/itex]
    So just the distance itself changes the resulting velocity. But velocity is supposed to be constant.
     
  8. Oct 2, 2015 #7
    Hmm...I'm looking at the coefficient of kinetic friction, mass of the object, and applied force of 5N. The only additional information I can gather from this is that the acceleration is zero. I can't do anything with net force because net force being zero turns any formula to zero. Kinematic equations don't seem to work because I don't have position, time, or velocity values. Even when I consider velocity to be equal to initial velocity or even final velocity, the fact that I don't have any information about distance or time, I can't solve for velocity in a kinematic equation.
     
  9. Oct 2, 2015 #8

    gneill

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    Zero is a perfectly good value :wink:
     
  10. Oct 2, 2015 #9

    [itex] F*d = \frac{1}{2}mv^{2} - \frac{1}{2}mv^{2} [/itex]

    [itex] 0N*d = \frac{1}{2}(5kg)v^{2} - \frac{1}{2}(5kg)v^{2} [/itex]

    [itex] 0 = \frac{1}{2}(5kg)v^{2} - \frac{1}{2}(5kg)v^{2} [/itex]

    [itex] \frac{1}{2}(5kg)v^{2} = \frac{1}{2}(5kg)v^{2} [/itex]

    [itex] v = v[/itex] ?
     
  11. Oct 2, 2015 #10

    gneill

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    Unless you're given the initial velocity it's not worth pursuing any calculations if the net force acting is zero. No change in the current state can happen. In fact, you were not given a time interval either. The only thing you can conclude is that the velocity remains unchanged from some (unknown) initial value.
     
  12. Oct 2, 2015 #11
    Suppose I push the block with more than 5N, say 10N. The net force will be 5N. The acceleration will be Fnet/m = 5N/5kg = 1m/s^2 = a.

    If I push the block with 10N of force for 10 seconds, then it can at least be determined that the speed of the block has increased 10m/s regardless of its initial velocity.

    For instance:

    If initial velocity is 0m/s:
    [itex] vf = at + v0 [/itex] = [itex] vf = (1m/s^2)(10s) + 0 = 10m/s [/itex]

    If the initial velocity is 5m/s:
    [itex] vf = at + v0 [/itex] = [itex] vf = (1m/s^2)(10s) + 5m/s = 15m/s [/itex]

    So the acceleration of 1m/s^2 for 10 seconds always produces an increase in velocity of 10m/s from whichever speed it started at.

    So an applied force of 6N can produce many different velocities depending on how much time the force of 6N is applied and the initial velocity.

    However, it seems as though 5N will produce only 1 specific velocity, regardless of how much time 5N is applied because acceleration will be 0.

    However the velocity might be different depending on the initial velocity. Previously, I was assuming there could only be one value for initial velocity, but now I realize that initial velocity can be any value. Thank you.

    So now my thinking is like this, for example:

    Suppose the 5kg block is at rest on the table with μk = .1 and for simplicity, suppose there is no static friction.

    So if I apply a force ever so gently to the block and increase the applied force by 1N every 10 seconds, the block will accelerate. And once the applied force of 5N is reached, the block will begin to travel at a constant velocity. Lets call it V1.

    Then, I make the block stationary again. I apply a force and increase the force at a faster rate, perhaps increasing by 1N every 5 seconds. The block will accelerate over this time and once I reach an applied force of 5N, the block will begin to travel at a constant velocity, but not the same constant velocity as V1, perhaps V2.

    To stress the example, if the 5kg block starts at rest again, and I apply 1N for 100 seconds, then jump all the way to 5N in 1 second, then the block will be traveling at a constant velocity but at a different velocity than V1 and V2.

    Would this be correct?
     
  13. Oct 2, 2015 #12

    gneill

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    Excellent so far!
    The thing about kinetic friction is that it is constant so long as the block is moving, no matter the velocity. Postulating that there is no static friction when kinetic friction exists is problematical for the case where the block is not moving when a force less than the friction is first applied. In fact, for most real-life situations the coefficient of static friction almost always exceeds that of kinetic friction. Perhaps a better approach would be to presume that μs = μk.
    As you say, over time nearly any velocity is possible depending upon the force applied. In the current problem There is not enough information given to reach a specific value other than it will not differ from the initial velocity (even if it's 0 m/s).

    What sort of answer is expected for this problem? Is it multiple choice, numerical, text, essay, something else?
     
  14. Oct 2, 2015 #13
    Thank you gneill,

    That makes sense. I'll definitely include static friction and for simplification, presume μs = μk. Plus, I just noticed, if 5N of applied force is net force = 0, then any applied force less than 5N would mean probably not even cause motion because the Net force would be in the opposite direction. Applied [ Force - Friction Force = Net Force] = [4N - 5N = -1N].

    I made up the question to continue learning about and understanding force, so maybe it's all of the above? :) And your advice on this question has lead to a new question: How to find the final velocity of an object given varying magnitudes of applied force over a period of time.

    I wonder if by finding out the final velocity of an object after a force that changes over time is applied, then perhaps I will be able to determine the velocity of an object when it reaches equilibrium by making the final force applied be 5N.

    Since this new question differs a bit from the original topic question for this thread, I'll post it as a new thread.

    Thank you, gneill. Appreciate your guidance on these topics.
     
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