# Calculate work based on resistive force

1. Sep 26, 2009

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

A bicyclist rides 5.7 km due east, while the resistive force from the air has a magnitude of 2.8 N and points due west. The rider then turns around and rides 4.6 km due west. The resistive force from the air on the return trip has a magnitude of 1.9 N and point due east.
(a) Find the work done by the resistive force during the round trip.

2. Relevant equations

W= F * D

3. The attempt at a solution
2.8N+1.9N (4600m + 5700m) = wrong answer!!

2. Sep 26, 2009

### Pengwuino

What you wrote as a solution is certainly not F * D. You basically tried to calculate F + F*D. You should have immediately realized this was wrong since you can't add 2 values of different units. Break the system into 2 parts, $$F_1$$ being the force on the rider and $$D_1$$ being the distance the rider travels subject to that force. Do the same for the trip back with $$F_2$$ and $$D_2$$. Your work is now $$W = F*D$$ which can be broken up into $$W = F_1 * D_1 + F_2 * D_2$$.

3. Sep 26, 2009

Pengwuino,
you said adding the two forces are wrong because they are different units...but they are both newtons, doesn't that mean its ok to add the forces together?
2.8(5700) +1.9(4600) = 24700
This is still wrong :(

4. Sep 26, 2009

### Pengwuino

Yes, but what you did was say $$W = F_1 + F_2(D_1 + D_2)$$ which was adding a force to a unit of energy.

I should have mentioned this the first time, but also remember that since the air took energy away from the rider, the work the air did (which is what's being asked for) is negative.

5. Sep 26, 2009

Ahhh. thank you very much that works out. However, at the risk of sounding stupid, I am still confused. Since both amounts are resistive forces, how is it that F1 + F2 is adding a force to a unit of energy?
Energy = joules?

6. Sep 26, 2009

### Pengwuino

When you write out $$F_1 + F_2(D_1 + D_2)$$, mathematical order of operation dictates you first add up $$D_1 + D_2$$ and then multiply them against $$F_2$$ which gives you units of energy. THEN you add $$F_1$$ but this is not valid since you're adding a force to an energy (which yes, is in Joules).

7. Sep 26, 2009