How much work is done on the box by person pulling?

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Homework Help Overview

The discussion revolves around a physics problem involving work done on a box being pulled at an angle, taking into account friction and various forces acting on the box. The problem includes calculations for work done by different forces and seeks to understand the relationships between them.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the pulling force, normal force, and friction. There is a specific focus on identifying errors in the calculation of work done by friction, with attempts to clarify the formula and approach used.

Discussion Status

The discussion is ongoing, with one participant expressing confusion about their calculation for the work done by friction and seeking clarification. Others are attempting to provide insights into the correct approach for this calculation.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available or the methods they can use. There is an emphasis on understanding the forces involved and their contributions to the work done on the box.

Bones
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Homework Statement



In the figure below,

http://www.webassign.net/userimages/jshemwell@lincolnpark.il/Net%20Force/Pulling_at_angle.gif

the mass is 4.00 kg, and = 30 degrees. The surface does have friction, with s = .45 and k = .39.

If a person takes hold of the rope and, pulling at this angle with a force of T = 29 N, drags the box 3.3 meters...

a. How much work is done on the box by person pulling?

b. How much work is done on the box by the normal force?

c. How much work is done on the box by the force of friction?

d. What is the net work done on the box?

e. If the block starts from rest, what will be its velocity just as the person finishes pulling?

Homework Equations



The Attempt at a Solution



a) 29N*3.3m*cos30 = 82.9 J
b) 0
c) I took 4kg*9.8m/s^2*0.39*3.3m*cos180 = -50.45 which is not correct. What am I doing wrong??
 
Last edited:
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I can figure out the rest, I just need to know what I am doing wrong in part c. I am stuck on part c.
 
Bones said:
c. How much work is done on the box by the force of friction?

The force of friction over the distance.

W = F * d = u*(m*g - T*sinθ ) * d
 
Thanks again!
 

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