How Do Applied and Frictional Forces Compare in a Constant Speed Scenario?

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SUMMARY

The discussion focuses on the relationship between applied and frictional forces in a scenario where a toy duck moves at a constant speed. It concludes that the applied force must be greater than the frictional force for the duck to waddle, as friction is the force that propels it forward. With a coefficient of friction of 0.15 and a mass of 10 kg, the maximum forward acceleration calculated is 1.47 m/s², aligning with textbook values. The analysis confirms that the friction force is essential for motion, and the acceleration is consistent regardless of mass, as it is derived from the frictional force equation.

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  • Understanding of Newton's laws of motion
  • Familiarity with the concept of friction and its coefficient
  • Basic knowledge of kinematics equations
  • Ability to perform algebraic manipulations for physics equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion, particularly in relation to friction and applied forces.

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Homework Statement


A toy duck waddles across the floor at constant speed.
1) How do the magnitubes of Fapplied and Ffriction compare?
2) If the coefficient of friction is 0.15, what is the maximum forward accelereation the duck can give itself?

Homework Equations


Ff = coefficient of friction x Normal Force
Fnet = ma
Fg = mg

Kinematics equations:
v2^2 = V1^2 + 2ad
V2 = V1 +at

The Attempt at a Solution


1) There are two answers i have considered. The first is that there is no applied force done to the object, because the duck is moving of its own accord. The second is that the applied force is equal to the force spent in winding up the duck and that that force is larger than the force of friction.
2)
I tried substituting 10 kg for the m.
So, Fg = 9.8 x 10 =98N
Then, Fn = 98N
So Ff = 98 x 0.15 = 14.7N
Fnet = ma
14.7 = 10 x a
a = 1.47 m/s^2.
This is the same as the textbook answer, but I'm not sure if i went about it the right way, nor am i sure about how this answer is right algebraically. Should Ff be equal to Fnet?
Thanks for the help :)
 
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1010 said:

Homework Statement


A toy duck waddles across the floor at constant speed.
1) How do the magnitubes of Fapplied and Ffriction compare?
2) If the coefficient of friction is 0.15, what is the maximum forward accelereation the duck can give itself?

Homework Equations


Ff = coefficient of friction x Normal Force
Fnet = ma
Fg = mg

Kinematics equations:
v2^2 = V1^2 + 2ad
V2 = V1 +at

The Attempt at a Solution


1) There are two answers i have considered. The first is that there is no applied force done to the object, because the duck is moving of its own accord. The second is that the applied force is equal to the force spent in winding up the duck and that that force is larger than the force of friction.
2)
I tried substituting 10 kg for the m.
So, Fg = 9.8 x 10 =98N
Then, Fn = 98N
So Ff = 98 x 0.15 = 14.7N
Fnet = ma
14.7 = 10 x a
a = 1.47 m/s^2.
This is the same as the textbook answer, but I'm not sure if i went about it the right way, nor am i sure about how this answer is right algebraically. Should Ff be equal to Fnet?
Thanks for the help :)
You have the correct equation (F_net = ma), but it is not consistent with either of the two considerations you noted. There must be an applied force in order for the duck to move. This force is provided by the friction force between the floor and its feet, whether its a toy duck or a real live one. Unless it was rocket powered, simply winding it up would not allow it to waddle if there was no friction under its feet. There must be a contact force to get it moving. So it is the friction force that actually propels the duck. The max value of the friction is as you noted for a 10Kg mass, but it doesn't matter what the mass is, the acceleration would be the same(the masses cancel out), the friction force is just 0.15mg, the acceleration is just .15g. Now the duck can apply less friction than the max, it would just waddle slower, like if the battery got weak.
 

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