Finding the net work done on an object

  • Thread starter Shyamalan
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Homework Statement


A 325 N force acts on an object with a mass of 42.5 kg at an angle of 27° , pulling the object along a flat surface. The coefficient of friction is 0.21 and the object's displacement is 320m.
Calculate the net work done on the object.

Homework Equations


W=Fdcos(θ) Fg=mg Ffr=μFN


The Attempt at a Solution


Ok, so I know FN in this case is equal to Fg, and the work done by the two is 0 J, because they perpendicular to the displacement. The work done by the pulling force is 325*320*cos(27) which is 92665 J. The friction force is 0.21*42.5*9.8 which is 87 N. The work done by the friction force is 87*320*cos(180) which is -27840 J. The net work is the sum of these two values, 92665+(-27840) which is 64825 J.

Now, the answer given is 7.46*10^4 J, which is one digit off of my answer. Am I doing something wrong or is there a typo in the answer I was given?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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Welcome to PF!

Hi Shyamalan! Welcome to PF! :smile:
Ok, so I know FN in this case is equal to Fg …

Noooo :rolleyes:
 
  • #3
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Pfffft... Whoopsy, I got it now. XP
 

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