# Work done by air resistance

1. Feb 8, 2007

### endeavor

1. The problem statement, all variables and given/known data
An airplane pilot fell 370 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot's mass was 80 kg and his terminal velocity was 50 m/s, estimate: (a) the work done by the snow in bringing him to rest; (b) the average force exerted on him by the snow to stop him; and (c) the work done on him by air resistance as he fell.

I need help on part c.
2. Relevant equations
$$W = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$$

3. The attempt at a solution

taking v2 = 50 m/s and v1 = 0, the work would be 100000 J. But shouldn't the work done be negative, because the force of air resistance is upward and the displacement is downward?

2. Feb 9, 2007

### andrevdh

You can attempt part a with the work - kinetic energy theorem, or could you do it by yourself.

3. Feb 9, 2007

### andrevdh

Yes, the work done by the friction is (always) negative, but the final kinetic energy is zero. You also need to include the work done by the weight of the falling man.

4. Feb 9, 2007

### PhanthomJay

In part c, your formula calculates the the total work done during the fall by both gravity (a conservative force) and air resistance (a non-conservative force). You are looking for the work done by the non-conservative air resistance force alone. Refamiliarize yourself with the work energy equations, which can get a bit tricky: Total work is the change in KE; Work by conservative forces is the negative of the change in PE; work done by non conservative forces is the change in KE plus the change in PE.

5. Feb 9, 2007

### andrevdh

For part (c) you would say that the work done by drag on the pilot is equal to the change in his mechanical energy. The mechanical energy of the pilot is his kinetic and potential energy combined. When the pilot reaches the ground his potential energy is zero and his kinetic energy is calculated with the 50 m/s terminal speed.

6. Feb 10, 2007

### endeavor

Someone told me that the work done by the air resistance would be the kinetic energy of the man if he did not reach a terminal velocity minus the kinetic energy of the man when he's at terminal velocity. So that would be:
$$W_a = \frac{1}{2}m(2gh) - \frac{1}{2}mv_t^2$$ with h = 370m.

does this work?

(Also, this problem is from the chapter before potential energy)

7. Feb 10, 2007

### denverdoc

He's right and the first term actually is the potential energy at the start of the fall. U=mhg=1/2mVf^2 + Wd, which is what Andrevdh suggested above, where Vf=term. vel.

8. Feb 10, 2007

### AngeloG

For part A, you have it some-what correct. The way I see it (and it is).

Ws + Ww = (DELTA) KE = 1/2mv^2 - 1/2mv^2

Where v prime = 0 m/s, v = 50 m/s; since v = 0. Ws = Work snow, Ww = Work weight.

Ws + Ww = - 1/2mv^2;
Ws = -1/2mv^2 - Ww

Where Ww = mgd;

Ws = -1/2mv^2 - mgd;
Ws = -(0.5)(80)(50^2) - (80)(9.80)(1.1)
Ws = -1 x 10^5 J.

However, that is the work done by the snow bringing him to a rest. Find it for the air resistance -- that too should be negative.