# Homework Help: Calculating work to overcome drag?

1. Nov 12, 2009

### sirfinklstin

1. The problem statement, all variables and given/known data
A blimp of mass 110 kg is pulled at an angle theta = 52 degrees downwards with respect to the horizontal for D = 7 km on level ground at a constant velocity v = 14 m/s. If the coefficient of drag (K in F = Kv^2) is 0.5 kg/m, how much work is done by the person pulling?

As I see this problem, the only thing that is relevant is drag, becuase there is no acceleration and the buoyancy of the blimp (not described) counter-acts gravity.

I calculated K in F and got 2112880, with v^2 as 196, and K as 10780 using (1/2)mv^2.
This does not seem right, help?

2. Nov 12, 2009

### cepheid

Staff Emeritus
Where do you get the 10780 from?

EDIT: You got it by calculating the kinetic energy of the blimp. But this is silly because K is not kinetic energy! K is the drag coefficient, whose value is given to you in the problem! You should stop and think whether your solution method makes any sense before proceeding with it. Why would the kinetic energy appear in this context?

Last edited: Nov 12, 2009
3. Nov 16, 2009

### sirfinklstin

So the work would be 98 joules? that doesn't seem right either.

4. Nov 16, 2009

### cepheid

Staff Emeritus
I get that the magnitude of the drag force is 98 N. That's not equal to the work done, though. I think you are missing something.

5. Nov 18, 2009

### sirfinklstin

I am completely stuck, becuase I cannot use the work-energy theorem becuase there is no acceleration (F = ma), so I cannot calculate anything! Please help!

6. Nov 18, 2009

### cepheid

Staff Emeritus
What do you mean? You are given the acceleration:

So you should know exactly what the net force is.

7. Nov 19, 2009

### sirfinklstin

Yes, at a constant velocity, which means acceleration is 0. I was told by my physics teacher to add the air resistance to the answer of the work energy theorem. Maybe (1/2)mv^2? Sorry, I seem to be caught in a web of stupidity on this one, and this problem is long overdue.

8. Nov 19, 2009

### cepheid

Staff Emeritus
The point is, that if the acceleration is zero, the *net* force on the blimp is zero (by Newton's Second Law). Therefore, whatever force is being used to pull on the blimp is only just barely enough to counteract the drag force on the blimp (the two forces are equal in magnitude and opposite in direction, hence they add up to zero). Therefore, if you calculate what the drag force is from F = Kv^2, then you know what the pulling force is too.

So now you are in a position to answer the question, "how much work is done by the person pulling?" To answer this question, you need to multiply the pulling force by the distance travelled. Be careful -- some trigonometry is involved in determining the actual distance through the air travelled by the blimp (if I am interpreting the problem wording correctly).

Just a clarification: even though F*d work is done on the blimp by the person pulling it, it remains at a constant velocity. This is not a contradiction of the work-energy theorem. Remember that the net force is zero, so the total work done is zero. The drag force does the same *negative* work on the blimp as the pulling force does positive work.

9. Dec 1, 2009

### sirfinklstin

Ok, I calculated a velocity of 5.26 m/s, but can someone please define the initial velocity and final velocity?????????????????????
Is initial velocity as soon as the ball drops or as soon as it hits the ground???

10. Dec 9, 2009

### cepheid

Staff Emeritus
I don't think there is a ball dropping in this problem. Are you sure you posted in the right thread?