# Work Done By Friction

1. Nov 25, 2004

### Jatsix30

I have no clue how to do this problem :

A flight attendant pulls her 75.2 N flight bag a distance of 268 m along a level airport floor at a constant velocity. The force she exerts is 33.0 N at an angle of 69.9 degrees above the horizontal. Find the work done by the force of friction on the flight bag. Answer in units of J.

So far, I figured that the total the force parallel to the surface is 11.34 N...

2. Nov 25, 2004

### ponjavic

Ok I don't know what you mean by the parallel force (as I am a stupid Swede)
Think about what friction is, the force of friction equals what?
(I forgot the letters) F(friction)=mu(friction coeffecient)*Fn(the force from the ground repelling the bag) damn i need to work on my English... anywho
So in order to calculate the friction force, which you will use to calculate the friction work since work = force*distance you will need to figure out Fn (assuming that you know the friction coeffecient).
If you split up the force she exerts into x and y axises you can get the force she exerts upwards.
What must the result of all forces working in the y axis be since the bag is only moving horisontally? From here you should be able to get Fn thus friction force thus friction work.
If you don't understand something I'll gladly explain, elaborate
I hope you understand that I do not want to give you a straight answer

3. Nov 25, 2004

### Chen

You're right, the X component of the pulling force is 11.34 N. And since the attendant is pulling her bag at a constant velocity, the sum of all forces in the X axis (and Y, but we don't care about that) must be 0. Therefore:

$$\Sigma F_x = F_{pull}\cos \alpha - f_k = 0$$

So f(k) = 11.34 N. And what's the work of that force over a distance of 268 meters?

$$W = \int _0^{268m}{\vec{f_k}\cdot d\vec{r}}$$

Just remebmer that f(k) is always in the opposite direction of dr, so the dot product will be -f(k)dr, and the work will be negative. Final answer should be 3039 J.

4. Nov 25, 2004

### futb0l

hmmm..............

5. Nov 25, 2004

### futb0l

what about the 75.2 N flight bag - since u can get the friction force parallel to the surface: (mg = 75.2N)

$$F_{friction} = mg \sin \alpha$$

6. Nov 25, 2004

### Jatsix30

Thank you both so much.

btw, the answer was in fact -3039

7. Nov 25, 2004

### futb0l

so the 75.2 N put on the question does nothing...

8. Nov 25, 2004

### futb0l

nevermind - i get it now :)

9. Nov 26, 2004

### Chen

I can't really understand why the included the bag's weight in the problem, unless one of the following questions would be "What's the coefficient of kinetic friction between the bag and the floor".

10. Nov 30, 2004

### inportb

That, by the way, was a problem found in the problem bank of hw.utexas.edu. I assume that the topic starter is a Stuy kid in SPXR, so a solution involving integration was probably not necessary .

The weight of the bag tells you that the bag is, in fact, travelling parallel to the ground and is not being lifted off the ground . Of course, the weight of the bag is offset by the y-pull.

Chen, you're right. The problem comes in three parts (my question had different numbers):

A flight attendant pulls her 62.2 N flight bag
a distance of 298 m along a level airport floor
at a constant velocity. The force she exerts is
38.1 N at an angle of 66.7° above the horizon-
tal.
a) Find the work she does on the flight bag.
Answer in units of J.
b) Find the work done by the force of friction
on the flight bag. Answer in units of J.
c) Find the coefficient of kinetic friction be-
tween the flight bag and the floor.