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

In summary, the conversation discusses a train of mass 8.0 x 10^6 kg, traveling at a speed of 30 m/s, that brakes and comes to rest with a constant deceleration in 25 seconds. The first question asks for the frictional force acting on the train while decelerating, which can be found using the equation F = ma and calculating the acceleration first. In the second question, the stopping distance of the train is requested, which can be calculated using the equation s = 1/2at^2 by determining the unknowns of acceleration and time.
  • #1
lina45
3
0

Homework Statement


Q : A train of mass 8.0 [tex]\times[/tex] 10[tex]^{}6[/tex] kg, traveling 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.


Homework Equations



I have no idea which equations to use...



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 don't know what do do next.
 
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  • #2
lina45 said:
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?
 
  • #3
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?
 
  • #4
but wouldn't the net force be zero because when you use the formula F = ma isn't constant deceleration regarded as zero acceleration?
 
  • #5
lina45 said:
but wouldn't the net force be zero because when you use the formula F = ma isn't constant deceleration regarded as zero acceleration?
Careful. Constant velocity is zero acceleration.
 
  • #6
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]
 
  • #7
lina45 said:
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:

1. What is frictional force?

Frictional force is the force that opposes the motion of an object when it comes into contact with another surface.

2. How is frictional force calculated?

Frictional force is calculated by multiplying the coefficient of friction (a constant value dependent on the two surfaces in contact) by the normal force (the force perpendicular to the surface).

3. How do you measure the deceleration of an object?

Deceleration can be measured by calculating the change in velocity over a certain period of time. This can be done using the formula: a = (vf - vi)/t, where a is the deceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

4. What factors affect the frictional force acting on an object?

The factors that affect frictional force include the type of surfaces in contact, the force pressing the surfaces together, and the roughness of the surfaces.

5. How can you reduce frictional force on an object?

Frictional force can be reduced by using lubricants between surfaces, using smoother surfaces, reducing the force pressing the surfaces together, or using wheels or ball bearings to reduce the surface area in contact.

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