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How do you find the frictional force acting on an object if it is decelerating?

  • Thread starter lina45
  • Start date
1. The problem statement, all variables and given/known data
Q : A train of mass 8.0 [tex]\times[/tex] 10[tex]^{}6[/tex] kg, travelling at a speed of 30 m s[tex]^{}-1[/tex], brakes and comes to rest with a constant deceleration in 25s.

(a) Calculate the frictional force acting on the train while decelerating.
(b) Calculate the stopping distance of the train.


2. Relevant equations

I have no idea which equations to use...



3. The attempt at a solution
Well... first i found the net force which is zero because it is contant negative acceleration. I then attempted a diagram. I dont know what do do next.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 
2,036
2
Well... first i found the net force which is zero because it is contant negative acceleration.
Are you sure? Would Newton's second law agree with that statement?
 

CompuChip

Science Advisor
Homework Helper
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In the first question, you are asked for a force. I know an equation that gives you the net force, namely: F = m a. But, there is an unknown in it, the acceleration. Since you know the initial velocity and stopping time, can you calculate the acceleration from this? And once you have the net force, is this also the force requested or might this be a sum of the force you actually want and some other force(s)?

In b, you need an equation that will give you the distance. Try [tex]s = \tfrac12 a t^2[/tex] - which unknowns are there and how can you determine them?
 
but wouldnt the net force be zero because when you use the formula F = ma isnt constant deceleration regarded as zero acceleration?
 
but wouldnt the net force be zero because when you use the formula F = ma isnt constant deceleration regarded as zero acceleration?
Careful. Constant velocity is zero acceleration.
 
howcome the formula s = [tex]\frac{}{}1/2[/tex] at[tex]^{}2[/tex] used instead of s = ut + [tex]\frac{}{}1/2[/tex] at[tex]^{}2[/tex]
 
howcome the formula s = [tex]\frac{}{}1/2[/tex] at[tex]^{}2[/tex] used instead of s = ut + [tex]\frac{}{}1/2[/tex] at[tex]^{}2[/tex]
Using s to denote displacement and u to denote velocity, the full kinematics equation for displacement is [tex]s = s_0 + u_0 t + \frac{1}{2} at^2[/tex], where [itex]s_0[/itex] is initial displacement and [itex]u_0[/itex] is initial velocity. If both of these are zero, they just drop out. So that just leaves us with [tex]s = \frac{1}{2} at^2[/tex]. Hope this helps.
 
Last edited:

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