Calculating Work for a Pulled Carton Across Floor

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Homework Help Overview

The problem involves calculating the work done by a horizontal applied force on a 21 kg carton being pulled across a floor at constant speed, considering the coefficient of kinetic friction. The context includes concepts of work, energy, and forces acting on the carton.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of constant speed on net force and question how to determine the normal force in relation to kinetic friction. There is an exploration of the relationship between applied force and friction force.

Discussion Status

Some participants have provided clarifications regarding the determination of normal force and its relationship to the applied force and friction. There is an ongoing exploration of the concepts involved, with hints being offered to guide understanding.

Contextual Notes

Participants note the importance of understanding the forces acting on the carton, including the effects of acceleration and the setup of the problem in a horizontal context.

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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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