How Is Work Calculated for a Refrigerator Moving Across a Surface?

AI Thread Summary
The discussion focuses on calculating the work done by the kinetic frictional force and the pulling force on a refrigerator being moved across a surface. The participant initially struggles with the calculations, yielding a work value of 1130 J for friction, which does not match the book's answer of -1200 J. After clarifying the angle between the friction force and displacement as 180 degrees, the correct work done by friction is confirmed to be -1200 J. The work done by the pulling force is calculated to be 1800 J. The conversation emphasizes the importance of correctly identifying angles in work calculations.
jahrollins
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Homework Statement


2.40 *102 N force is pulling 85.0 kg refrigerator across a horizontal surface. The force acts at an angle of 20.0 degrees above the surface. The coefficient of kinetic friction is 0.200 and the fridge moves a distance of 8.00 m. Find the the work done by the kinetic frictional force and the pulling force.


Homework Equations


Equation 1
W = F(cos(angle))*s
s = displacement

Equation 2
Fs = Fn * coefficient of kinetic friction



The Attempt at a Solution


So I'm just working to get the work done by the kinetic frictional force first. My answer doesn't match the book's though.
Fn - mg + 240sin20= m*ay
Fn - mg + 240sin20 = 0
Fn = mg - 240sin20

Put Fn = mg into equation 2.
Fs = Fn * coefficient of kinetic friction
Fs = (mg - 240sin20) * coefficient of kinetic friction

Put into equation 1
W = F(cos(angle))*s
W = (mg - 240sin20) * coefficient of kinetic friction(cos(angle))*s

Solve
W = (85kg*9.8m/s2 - 240sin20) * 0.200(cos(20))*8m
W = 1130 J
Book says -1.2*103 J
 
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jahrollins said:

Homework Statement


2.40 *102 N force is pulling 85.0 kg refrigerator across a horizontal surface. The force acts at an angle of 20.0 degrees above the surface. The coefficient of kinetic friction is 0.200 and the fridge moves a distance of 8.00 m. Find the the work done by the kinetic frictional force and the pulling force.


Homework Equations


Equation 1
W = F(cos(angle))*s
s = displacement

Equation 2
Fs = Fn * coefficient of kinetic friction



The Attempt at a Solution


So I'm just working to get the work done by the kinetic frictional force first. My answer doesn't match the book's though.
Fn - mg + 240sin20= m*ay
Fn - mg + 240sin20 = 0
Fn = mg - 240sin20

Put Fn = mg into equation 2.
Fs = Fn * coefficient of kinetic friction
Fs = (mg - 240sin20) * coefficient of kinetic friction

Put into equation 1
W = F(cos(angle))*s
W = (mg - 240sin20) * coefficient of kinetic friction(cos(angle))*s
what is the angle beween the friction force and the displacement?
Solve
W = (85kg*9.8m/s2 - 240sin20) * 0.200(cos(20))*8m
W = 1130 J
Book says -1.2*103 J
correct for the proper angle between the friction force and displacement, and your answer for the work done by friction will be correct (watch plus /minus signs, however). Then also find the work done by the applied pulling force (part 2 of the question).
 
Ah, okay. Then since the the frictional force would be opposite the displacement the angle would be 180? That makes it
W = (85kg*9.8m/s2 - 240sin20) * 0.200(cos(180))*8m
W = -1200 J
 
jahrollins said:
Ah, okay. Then since the the frictional force would be opposite the displacement the angle would be 180? That makes it
W = (85kg*9.8m/s2 - 240sin20) * 0.200(cos(180))*8m
W = -1200 J
Yes..now on to part 2...if you would, please.
 
So, basically just solve since we have all the values already...
W = 240N(cos(20))*8m
W = 1800J
 
jahrollins said:
So, basically just solve since we have all the values already...
W = 240N(cos(20))*8m
W = 1800J
:cool:Now you're making it look easy...:smile:
 

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