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

[jahrollins]

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

 

[PhanthomJay]

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).

 

[jahrollins]

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

 

[PhanthomJay]

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.

 

[jahrollins]

So, basically just solve since we have all the values already...
W = 240N(cos(20))*8m
W = 1800J

 

[PhanthomJay]

So, basically just solve since we have all the values already...
W = 240N(cos(20))*8m
W = 1800J

Now you're making it look easy...