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

[lina45]

Homework Statement

Q : A train of mass 8.0 \times 10^{}6 kg, traveling at a speed of 30 m s^{}-1, 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.

 

[neutrino]

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]

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 s = \tfrac12 a t^2 - which unknowns are there and how can you determine them?

 

[lina45]

but wouldn't the net force be zero because when you use the formula F = ma isn't constant deceleration regarded as zero acceleration?

 

[z-component]

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.

 

[lina45]

howcome the formula s = \frac{}{}1/2 at^{}2 used instead of s = ut + \frac{}{}1/2 at^{}2

 

[z-component]

howcome the formula s = \frac{}{}1/2 at^{}2 used instead of s = ut + \frac{}{}1/2 at^{}2

Using s to denote displacement and u to denote velocity, the full kinematics equation for displacement is s = s_0 + u_0 t + \frac{1}{2} at^2, where s_0 is initial displacement and u_0 is initial velocity. If both of these are zero, they just drop out. So that just leaves us with s = \frac{1}{2} at^2. Hope this helps.