Stuck on sliding object problem

In summary: You are correct, the coefficient of friction is needed to solve for the kinetic friction force. In this case, the coefficient can be found by dividing the kinetic friction force by the normal reaction force (mg). Therefore, the coefficient of friction is approximately 0.4.
  • #1
redshift
53
0
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?
 
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  • #2
redshift said:
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
 
  • #3
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.
 
  • #4
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)
 
  • #5
cipher said:
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)
 
  • #6
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.
 
  • #7
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
 

1. What is the "stuck on sliding object problem"?

The "stuck on sliding object problem" is a common physics problem that involves a mass being placed on a frictionless surface and then attached to a spring. The mass is then pulled back and released, causing it to slide and oscillate back and forth due to the spring's force. The problem arises when the mass gets stuck on the surface and stops sliding, leading to the question of what factors contribute to this phenomenon.

2. How can the "stuck on sliding object problem" be solved?

To solve the "stuck on sliding object problem," one must take into consideration the various forces acting on the mass, including the spring force, gravity, and any external forces. These forces can be represented using equations and can be solved using mathematical methods such as Newton's laws of motion and energy conservation principles.

3. What are the potential causes of the mass getting stuck on the sliding surface?

The mass can get stuck on the sliding surface due to various reasons, including an insufficient initial force, an incorrect choice of spring constant, or an uneven surface that affects the motion of the mass. Another possible cause could be the presence of external forces, such as friction or air resistance, that impede the sliding motion of the mass.

4. How can the "stuck on sliding object problem" be prevented?

To prevent the mass from getting stuck on the sliding surface, one can make adjustments to the initial force applied, the choice of spring constant, and the surface on which the mass is placed. Additionally, reducing or eliminating external forces can also help prevent the mass from getting stuck during its motion.

5. What are some real-world applications of the "stuck on sliding object problem"?

The "stuck on sliding object problem" has practical applications in various fields, including engineering, physics, and mechanics. Understanding the factors that contribute to this problem can help engineers design and optimize systems that involve oscillating or sliding objects, such as shock absorbers, car suspensions, and pendulum clocks.

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