Calculate work based on resistive force

[agadag]

Homework Statement

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.

Homework Equations

W= F * D

The Attempt at a Solution

2.8N+1.9N (4600m + 5700m) = wrong answer!
please help!

 

[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.

 

[agadag]

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?
However I tried your method
2.8(5700) +1.9(4600) = 24700
This is still wrong :(

 

[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.

 

[agadag]

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?

 

[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).

 

[agadag]

ohhh...ok that makes sense.
Thanks again!