How Do You Calculate Kinetic Friction Force in a Sliding Object Problem?

[redshift]

I'd appreciate a tip here.
A 10 kg object on a horizontal, unsmooth table is pushed to the right so as to have an intial velocity of 20 m/s. It comes to rest 5 seconds later, during which time its acceleration (deceleration?) was constant.

I've already figured out the acceleration (4 m/s^2) and distance covered until coming to rest (50 m). However, I can't see how to get the kinetic friction force.
Would the initial force (ma = 40 N) include the kinetic friction?

 

[arildno]

I'd appreciate a tip here.
A 10 kg object on a horizontal, unsmooth table is pushed to the right so as to have an intial velocity of 20 m/s. It comes to rest 5 seconds later, during which time its acceleration (deceleration?) was constant.

I've already figured out the acceleration (4 m/s^2) and distance covered until coming to rest (50 m). However, I can't see how to get the kinetic friction force.
Would the initial force (ma = 40 N) include the kinetic friction?

1. "acceleration (deceleration?) "
Either tems work; acceleration is the "scientific" term which doesn't bother to have 2 different terms depending upon whether the object slows down or speeds up;
(This usage of "acceleration" is the one meant in Newton's 2 law)
whereas the couple acceleration/deceleration are the "lay" terms which insist upon having two different terms..

2.
"However, I can't see how to get the kinetic friction force.
Would the initial force (ma = 40 N) include the kinetic friction?"

What do you mean by this?
In particular, what do you mean by "initial force"?
If you by "initial force" means the force initially speeding up the object, you are wrong in setting it equal to 40N.
This is because you do not know the acceleration you had in the time speeding up to 20m/s, (neither do you know the time it took, either)
you only know the acceleration during the slowdown phase (4m/s^(2)).

But that acceleration is in its entirety produced by kinetic friction force!
(You have stopped pushing the block).
Hence, the kinetic friction force is 40N

 

[Doc Al]

I presume when the object is "let go" the intial pushing force is removed. So the only force contributing to the deceleration is friction. Apply F = ma.

 

[exequor]

don't you have to have the coefficient of friction to find the force that you want to. the F = (coefficient of friction) x (the normal reaction)

 

[arildno]

don't you have to have the coefficient of friction to find the force that you want to. the F = (coefficient of friction) x (the normal reaction)

Excellent question, cipher, since that is what you normally would need.

However, the text explicitly states that the object comes to rest after 5 seconds.
You can therefore solve for the unknown, constant acceleration by the equation:
20-a*5=0, which implies a=4m/s^(2).

This gives the kinetic friction force to be 40N, and you may DEDUCE the friction coefficient, f, to satisfy:
40=f*mg (in this example, f will then be approximately 0.4)

 

[redshift]

Yes, by "initial force" I meant that force that brought the object to its initial velocity of 20 m/s. I see now that I was mixing accelerations of the speedup and slowdown phases. Thanks for your patience. I'm slowly getting the hang of this.

 

[KnowledgeIsPower]

You're studying mechanics? At least that's what we call it over here.

Work out acceleration:
U = 20, T = 5, V = 0, a = ?
V = U + AT
Substitute and simplify to find a = -4m/s/s.

Resolve the particle horizontally.
F = ma
-FR = 10 x -4
FR = 40N

Resolve vertically:
R = 10G
R = 98N

Fr = (Mue)R
Fr/R = (Mue)
40/98 = (Mue)
(Mue) = .41
CF = .41