Calculating Work for a Pulled Carton Across Floor

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To calculate the work done by a horizontal force pulling a 21 kg carton across the floor at constant speed, the applied force must equal the friction force due to the coefficient of kinetic friction (0.41). Since there is no acceleration, the net force is zero, indicating that the applied force balances the friction force. The normal force on a horizontal surface is determined by the equation N = mg, where m is the mass and g is the acceleration due to gravity. If the carton were on an incline, the normal force would be calculated based on the forces acting perpendicular to the incline. Understanding these concepts is crucial for solving similar physics problems effectively.
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Homework Statement



A horizontal applied force, F, pulls a 21 kg carton across the floor at constant speed. If the coefficient of kinetic friction between the carton and the floor is 0.41, how much work does the applied force do in moving the carton 2.5 m?

Homework Equations



I'm supposed to be working with energy techniques, we just covered work/power/energy conservation/and so on

The Attempt at a Solution



I don't quite understand, it seems to me like the force is 0 given there is no acceleration.
 
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DiscreteCharm said:
I don't quite understand, it seems to me like the force is 0 given there is no acceleration.
The fact that there is no acceleration tells you that the net force is zero. Hint: Figure out what the friction force must be and then deduce what F is.
 
That gives way to the problem I have been having in working out a few other problems. I know it's a simple question, but how do I determine normal force (since I'm almost always given the coefficient of kinetic friction)? Is it simply mass times gravity (on a horiztonal surface, at least)?
 
Yes, in this case, the normal force is simply N = mg. In general, you figure it out by setting the sum of the vertical forces equal to zero. The only vertical forces acting on the carton are -mg and +N, so N - mg = 0, and thus N = mg. (Sometimes the applied force will have a vertical component that will affect the normal force.)

And if the carton were being pushed up an incline instead of on a horizontal surface, you'd figure out the normal force by setting the sum of the forces perpendicular to the incline equal to zero.
 
Thank you very much, that was extremely helpful.
 
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