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

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In a constant speed scenario, the applied force must be equal to the frictional force for the toy duck to maintain its motion. The friction force, calculated using the coefficient of friction (0.15) and the normal force, is essential for propulsion. The maximum forward acceleration the duck can achieve is determined by the friction force, which is 0.15 times the gravitational force acting on the duck. The calculations confirm that the acceleration is consistent regardless of mass, as the forces balance out. Thus, the friction force is crucial for the duck's movement and determines its maximum acceleration.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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