Work Done by Normal Force: 40kg Box Pushed 5m over Rough Floor

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Homework Help Overview

The problem involves a 40.0 kg box being pushed 5.00 m along a rough horizontal floor with a constant applied force of 130 N. The task is to determine the work done by the normal force, considering the coefficient of friction is 0.300.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the normal force, noting the relationship between force and displacement as vectors. There is an exploration of why the work done by the normal force is zero due to the angle between the force and displacement being 90 degrees.
  • Questions arise regarding the calculation of work done by friction and the interpretation of the normal force in relation to the weight of the box.

Discussion Status

Some participants provide clarifications regarding the calculations and concepts involved, particularly the relationship between the normal force and the weight of the box. There is an acknowledgment of understanding reached by some participants, but further questions remain about the frictional force and its calculation.

Contextual Notes

Participants are discussing the implications of the forces acting on the box, including the normal force and friction, within the context of Newton's laws. There is a focus on the free body diagram and the forces involved in the scenario.

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Homework Statement


A 40.0kg box initially at rest is pushed 5.00m along a rough, horizontal floor with a constant applied horizontal force of 130N. If the coefficient of friction between box and floor is 0.300, find the work done by the normal force?


Homework Equations


W=F*d


The Attempt at a Solution


The answer is 0N, however i got 1260(which i know does not seem logically correct).
So can someone explain to me why does work done by the normal force equal to 0N.
 
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because the work = F.d, don't forget that F and d (the force and displacment) are both vectors , F.d is a dot product, then you have work=F.d=Fdcostheta , where theta is the angle between the force(in your case is the normal force) and the displacement .. and since theta is 90 degree then w=Fdcos90=0J (the unit of work is J not N)...
 
thebigstar25 said:
because the work = F.d, don't forget that F and d (the force and displacment) are both vectors , F.d is a dot product, then you have work=F.d=Fdcostheta , where theta is the angle between the force(in your case is the normal force) and the displacement .. and since theta is 90 degree then w=Fdcos90=0J (the unit of work is J not N)...

ok thanks i understand, however another question i would like to ask, Then how would you explain how i got the work done by friction to be -588. Because to find the Frictional force i times the (coefficient of friction by the Normal Force ). However if you say that the Normal Force is mgcostheta then i don't understand why the answer would be -588, because W=-392cos90 * 5 then it would be 0 Joules?
 
who said that the normal force is mgcostheta? ...
If u carefully draw the freebody diagram u can see that u have 4 forces acting on the box ..
1- the force that horizontally applied which is 130 N ..
2- the force of friction which is opposing the force you applied and = coeffient of friction * normal force..
3-the weight of the box which is downward and = m*g ..
4-the normal force which is upward..

Sice the resulant motion is in the horizontal axes, then according to Newtons 2nd law .. The net force in the vertical axes = zero .. Which implies that : the normal force - the weight of the box = zero ..
Then, N = mg (and not mgcostheta) ...
So, the friction force = coeffient of friction* mg ..

Hopefully this was clear enough :) .. If u still have question ask again ...
 
yep thanks i understand now :)
 

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