Heat energy dissipated from moving box with friction

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Homework Help Overview

The problem involves calculating the rate of heat energy dissipation from a box being dragged across a rough surface, focusing on the effects of friction. The context includes the weight of the box, the speed at which it is dragged, and the coefficient of friction between the box and the floor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between power, force, and velocity, with one participant attempting to calculate the power based on the frictional force. There is mention of dimensional analysis and the suggestion to use a free body diagram for clarity.

Discussion Status

Participants are exploring the problem with some guidance provided regarding the use of free body diagrams and the relationship between force and power. There is acknowledgment of the calculations presented, but no consensus on the final answer has been reached.

Contextual Notes

There is a focus on understanding the forces acting on the box, including gravitational and frictional forces, and the implications of these forces on the calculations of power and energy dissipation. The discussion reflects on the assumptions made regarding the forces involved.

Pete Panopoulos
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Homework Statement


A box of books that weighs 40N is dragged at a speed of 1.5 m/s across a rough floor. If the coefficient of friction between the floor and the box is 0.20, what is the rate at which heat energy is dissipated?

Homework Equations


Friction: f=μN
Work: W=F*d
Power: P=W/t

The Attempt at a Solution


[/B]
My assumption would be if the units of power are watts, or kgm2/s3, breaking that up would be :

kg*m/s2*m/s

and the box weighs 40N, so kg*m/s2=40, and v=1.5m/s

With the coefficient of friction being 0.20,

would the answer be 40*1.5*0.2= 12W ?
 
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Pete Panopoulos said:

Homework Statement


A box of books that weighs 40N is dragged at a speed of 1.5 m/s across a rough floor. If the coefficient of friction between the floor and the box is 0.20, what is the rate at which heat energy is dissipated?

Homework Equations


Friction: f=μN
Work: W=F*d
Power: P=W/t

The Attempt at a Solution


[/B]
My assumption would be if the units of power are watts, or kgm2/s3, breaking that up would be :

kg*m/s2*m/s

and the box weighs 40N, so kg*m/s2=40, and v=1.5m/s

With the coefficient of friction being 0.20,

would the answer be 40*1.5*0.2= 12W ?

Your answer looks right, but rather than just doing the dimensional analysis, it would probably help you to draw a free body diagram. Later on, you will see vector equations for work where this dimensional analysis approach won't work.

For the free body diagram, you have gravity pushing down with a Force of 40N. This is compensated by a normal force of 40N (ground pushing up on the box). You know this, because there is no acceleration up/down.

There is also no acceleration forward-backwards. So the net force inthis direction is also zero. You are pushing the box forward with some force F->, friction is pushing back with an equal and opposite force <-F. The magnitude of the Frictional force is equal to the Normal Force times the coefficient of friction.

The work done is the force times the distance. The rate of energy lost to friction is equal to the work done divided by time.
 
I agree.

Work = force * distance
Power = work/time
so
Power = force * distance/time
or
Power = force * velocity.
 
Thank you very much!
 

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