How Much Work is Done When Pushing a Sled with Different Friction Forces?

[Senjai]

Homework Statement

A girl pushes her little brother on his sled with a force of 300N for 750m. How much work is done if the force of friction is a) 200N and b) 300N?

Homework Equations

W = F \Delta x

The Attempt at a Solution

F_{net} = F_p - F_f
F_{net} = 300N - 200N = 100N
W = F \Delta x
W = (100N)(750m) = 75000 N{\cdot}m = 7.5 x 10^4 J

That answer is wrong, its supposed to be 2.25 x 10^5 J

what did i do wrong?

 

[Gib Z]

Really the question should have stated what direction the work is to be computed in. Judging from what the answer is "supposed" to be, The Force it takes into account is 300N, not 100N. Basically what I think the question wanted you to think is that 300N is the resultant force perhaps? And they just gave you difference friction forces for fun? The question isn't worded very good.

 

[Senjai]

Yea it must be, as question a and b have the same answer, but if there was a force of friction, would you not have to exert more work on the object?

 

[Gib Z]

Yes you would, so I believe they missed the key word: Resultant.