# Average drag?

1. Mar 22, 2006

### D.T.

Average drag??

Hi' this is my first post here , What is average drag :uhh: (underlined bit) eg,
I shot a bullet thru three pieces of plasticene(15cm thickness each) each placed 5 cm apart and the average drag of the plasticene is 5000 N the bullet has a mass of 200g was travelling at 100m/s how far will the bullet go through the plasticane ?? :shy: I'd like to know how to work that out as well
thanks alot guys

2. Mar 23, 2006

### andrevdh

Drag is the average retarding force that an object (in this case the bullet) experiences as it moves through the medium (plasticine). The plasticine therefore generates friction that slows the bullet down as it moves through it. The drag force will do work on the bullet causing a reduction of its kinetic energy. If we ignore other forces acting on the bullet we can say that:
the work done on the bullet by the drag force acting on it = the change in the kinetic energy of the bullet
using the work-kinetic energy theorem.

3. Mar 23, 2006

### D.T.

Oh , I see now, thanks alot, can I use this formula to solve this problem??
E2 = E1 + W21 ??
0 = 1/2 MV2 + f.S ??

I don't know whether its a right formula or not please correct if I'm wrong
thanks again!!

4. Mar 23, 2006

### andrevdh

Yes, your formula can be rewritten as
$$EK_{final}-EK_{initial}=W_{drag}$$
where $EK$ is the kinetic energy of the bullet and $W_{drag}$ is the work done by the drag force on the bullet. That is the friction on the bullet is removing kinetic energy from the bullet as it moves throught the plasticene. The above equation reduces to
$$-EK_{initial}=W_{drag}$$
since the final kinetic energy of the bullet is zero - it came to a stop - as you showed in your equation above. What to do about the negative sign??!!

Well, the work done by the drag on the bullet is negative!

Last edited: Mar 23, 2006