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    The braking distance of a vehicle for a speed of 18ms^ and acceleration of 6.75ms^ on a dry level road is 24m. Calculate :
    the frictional force on a vehicle of mass 1000kg on this road when it stops?
    (how will you be able to calculate frictional force here?is it as simple as FORCE=MASS X ACCELERATION/DECELERATION? )
    if you multiply mass by deceleration, does it always give you the frictional force?
    Thanks for your help guys .
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    (Original post by Alen.m)
    The braking distance of a vehicle for a speed of 18ms^ and acceleration of 6.75ms^ on a dry level road is 24m. Calculate :
    the frictional force on a vehicle of mass 1000kg on this road when it stops?
    (how will you be able to calculate frictional force here?is it as simple as FORCE=MASS X ACCELERATION/DECELERATION? )
    if you multiply mass by deceleration, does it always give you the frictional force?
    Thanks for your help guys .
    As long as no other forces are acting on the object, then the braking force means frictional force in the context of this question.

    F = ma in all cases.

    But you need to be clear as to where the action/reaction is taking place.

    With this question, in the absence of any other information you can take it to mean the frictional force between the road wheels and the road.

    NB The deceleration of the vehicle is made up from the sum of all forces acting on the vehicle and there are several of these: For instance, road wheels with the road surface, brake pads with the road wheels, deceleration caused by drag (air resistance), incline of a hill (gravity) etc.

    Which means the final deceleration is the vector sum of all forces of which the braking force is just one.


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    thanks so much
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    (Original post by uberteknik)
    As long as no other forces are acting on the object, then the braking force means frictional force in the context of this question.

    F = ma in all cases.

    But you need to be clear as to where the action/reaction is taking place.

    With this question, in the absence of any other information you can take it to mean the frictional force between the road wheels and the road.

    NB The deceleration of the vehicle is made up from the sum of all forces acting on the vehicle and there are several of these: For instance, road wheels with the road surface, brake pads with the road wheels, deceleration caused by drag (air resistance), incline of a hill (gravity) etc.

    Which means the final deceleration is the vector sum of all forces of which the braking force is just one.


    Surely friction is the driving force in a car? Friction with the road would only apply a force backwards if you were braking and so the yellow arrow wouldn't exist for this case.
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    (Original post by lerjj)
    Surely friction is the driving force in a car? Friction with the road would only apply a force backwards if you were braking and so the yellow arrow wouldn't exist for this case.
    Yes, of course.

    I should have made it clear that the diagram is an example of the different forces acting on the car and was not meant to illustrate the braking force.

    Poor choice of diagram, thanks for pointing that out.
 
 
 
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