What is the frictional force at constant velocity?

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SUMMARY

The frictional force acting on a cart moving at constant velocity with a 10 N applied force is equal to 10 N, directed opposite to the applied force. This conclusion is derived from the understanding that when an object moves at constant velocity, the net force is zero, meaning all forces acting on the object must balance. The forces involved include the applied force, frictional force, weight, and normal reaction force, which cancel each other out accordingly.

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moonbase
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Homework Statement


A 10 N force is horizontally applied to a cart from the left. The cart moves to the right with a constant velocity. What is the magnitude and direction of the frictional force? (In terms of mass)

Homework Equations


Fnet=ma
fk=ukN

The Attempt at a Solution


I know Newton's second law shouldn't apply since the acceleration equals zero so there is no net force. But I'm confused as to how I can get all the forces to cancel out if the applied force needs to be greater than the frictional force in order for the cart to move. So the frictional force can't be 10 N. Is there a force I'm missing or is there something wrong with my reasoning?
 
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moonbase said:

Homework Statement


A 10 N force is horizontally applied to a cart from the left. The cart moves to the right with a constant velocity. What is the magnitude and direction of the frictional force? (In terms of mass)

Homework Equations


Fnet=ma
fk=ukN

The Attempt at a Solution


I know Newton's second law shouldn't apply since the acceleration equals zero so there is no net force. But I'm confused as to how I can get all the forces to cancel out if the applied force needs to be greater than the frictional force in order for the cart to move. So the frictional force can't be 10 N. Is there are force I'm missing or is there something wrong with my reasoning?

your problem lies in the statement in red, above.

You do, and have, used Newton's second law when you concluded "... so there is no net force".

getting all the forces to cancel out is easy ...

Weight down
Normal Reaction Force up
Applied Force to the right [applied from the left]
Friction Force to the left.

They cancel in pairs.
 
Oh okay, I was getting static and kinetic friction mixed up. Thanks
 

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