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

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AI Thread Summary
The discussion focuses on calculating the work done on a box being pulled at an angle, considering various forces involved. The person pulls with a force of 29 N at a 30-degree angle, and the box has a mass of 4.00 kg with friction coefficients of static and kinetic friction. The user successfully calculated the work done by the person pulling as 82.9 J but is struggling with the work done by friction. They are attempting to apply the correct formula for frictional work but are uncertain about their calculations. The conversation emphasizes the need to accurately account for the forces acting on the box to determine the net work done.
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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??
 
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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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