Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deceleration imposed by friction

  1. Jun 13, 2010 #1
    Ok, so I was doing this problem, *NOT* a homework problem, where I was solving for the force (and eventually, the work) required to move a body with a mass of m distance x in a given time y. I know that theoretically there would not be any force because there is no given acceleration and therefore no force ( F = mass * acceleration) but I considered that friction, etc. does not allow for perpetual movement at a constant speed, so with friction there has to be some acceleration to at the very least override the forces of friction and the like to have a net velocity > 0. Is there a fundamental equation that can be used to calculate the forces against a body? I know that deceleration due to gravity is -9.8(sin [tex]\Theta[/tex]) because gravity accelerates the falling of a body. At a 90 degree angle the deceleration is 9.8 m/s2. Any other slope follows that formula.

    I do have another question that is a bit off topic: Would it be fair to say that the more energy is used by a system, the more waste heat is generated and as a result q / w approaches 1?

    Keep in mind that my understanding of physics is not very broad thus far and everything I have learned is either intuitive classical physics I picked up through practical learning (like that gravitational formula), quantum physics taught by chemistry professors, and work examples given by calculus professors. I have never had a physics class in my life, including HS.

    Thanks everyone.
  2. jcsd
  3. Jun 13, 2010 #2

    jack action

    User Avatar
    Science Advisor
    Gold Member

    You don't need an acceleration to balance a force. For example you can pull a cable tie to a tree and you would have a force while nothing moves, hence no acceleration.

    Work equals the force times the distance traveled and the the power equals the work done divided by the time take to do that work.

    The friction force can be of two kind: Static friction and kinetic friction. Static friction is when you push on the object and it doesn't slide. In this case the friction force is always equal and opposite to the force that pushes on the object. Thus, the net force on the object is always zero, so no acceleration and no displacement. Since there is no displacement, there is no work done which leads to zero power too.

    But if the force is large enough, it will break a threshold point where the object starts to move. There will still be friction, but we call it kinetic friction as it is slightly weaker than the maximum static friction force. At this point, because there is a displacement, there is work done which equals the force that pushes the object (= friction force) times the distance traveled and we can get the power by dividing by the time. If the force that pushes the object becomes larger than the friction force, then the resultant force (= force that pushes - friction force) is not zero anymore and we can get the acceleration of the object by dividing this resultant force by the object's mass. The work done is equal to the forces that pushes the object (= friction force + mass*acceleration) times the distance traveled and so on.

    There is no direct relation between energy used and energy converted to heat. It all depends on the system. All we know is that energy is conserve. But it can be under the form of work, heat, potential energy (raising an object against gravity) and some other forms too, but the ratio can be anything depending on the system.
  4. Jun 13, 2010 #3


    User Avatar

    Staff: Mentor

    When you analyze a dynamic system in physics, you do a force balance:
    F1+F2 = 0
    For a non-accelerating moving object in a constant gravitational potential (not being driven uphill or downhill), all of the energy expended [almost] immediately becomes waste heat.
    Last edited: Jun 13, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook