# Homework Help: Work done by constant force

1. Feb 10, 2010

### jahrollins

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Equation 1
W = F(cos(angle))*s
s = displacement

Equation 2
Fs = Fn * coefficient of kinetic friction

3. 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

2. Feb 10, 2010

### PhanthomJay

what is the angle beween the friction force and the displacement?
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).

3. Feb 10, 2010

### 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

4. Feb 10, 2010

### PhanthomJay

Yes..now on to part 2...if you would, please.

5. Feb 10, 2010

### jahrollins

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

6. Feb 10, 2010

### PhanthomJay

Now you're making it look easy.........