Finding the net work done on an object

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To find the net work done on an object with a mass of 42.5 kg and a 325 N force acting at a 27° angle, the calculations involve determining the work done by the applied force and friction. The work done by the pulling force is calculated as 92665 J, while the friction force of 87 N results in -27840 J of work. The net work is then found by summing these values, resulting in 64825 J. However, there is a discrepancy with a provided answer of 7.46*10^4 J, prompting a discussion about potential errors in the calculations. The conversation highlights the importance of careful calculation in physics problems.
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?
 
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Welcome to PF!

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

Noooo :rolleyes:
 
Pfffft... Whoopsy, I got it now. XP
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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