# Work done by pulling a wagon.

1. May 6, 2007

### User1247

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

Verbatim from the text: "In many neighborhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50 kg and the adult does 2.2x10^3 J of work pulling the two 60 m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.

1.Determine the magnitude of the force applied by the parent.
2.Determine the angle at which the parent applies is applying the force."

2. Relevant equations
W=FΔd

3. The attempt at a solution
Note 1: Not sure what this vague thing about "the coefficient of friction" is, wagons usually have wheels and wheels usually have bearings, people don't usually pull things with the brakes on either so I guess we can just pretend that the wagon is wheel-less and has a kinetic friction coefficient of 0.26.

Fp=force applied by parent
Fpx=horizontal component of force applied by parent
Fg=force of gravity on wagon
Fn=normal force on wagon
Ff=frictional force on wagon
Wp=work done by parent on wagon
Mu=coefficient of friction

1. Find magnitude of Fp:
(first I solve for Fpx)

Find Fg, Fn, Ff:
Fg=mg=45kg x 9.8N/kg = 441N
Fn=441N
Ff=Fn x Mu
Ff=441N x 0.26
Ff=114.66N

Find Fnet:
Because the speed is constant, a=0
Fnet=ma
Fnet=m(0)
Fnet=0

Find Fpx
Fnetx=Ff+Fpx
0=-Ff+Fpx
Fpx=Ff
Fpx=114.66N

At this point I have a problem because it seems like what I have done makes sense so far, but with Fpx=114.66N we can find the work done by the parent, Wp = (Fpx)Δd = (114.66N)(60m) = 6879.6J
Which is more than the 2200J of work that the problem says the parent did.

If we start from a different route on solving the problem and say:
Parent does 2200J of work over 60m and we divide 2200J/60m to get the F by the parent || to the horizontal displacement we get
2200J/60m = 36.6666N Fpx

But now I have another problem: If the parent supplies 36.6666N of force, how can they overcome a friction force of 114.66N in the opposite direction and cause any displacement at all!?

Anyone have any ideas? I'm at a loss here . Perhaps it's a trick question and I'm to ignore the friction because the wagon has wheels? Even still, by working backwards through W=F||*displacement (where F|| is force parallel to displacement) there is no way I can see to get the actual Fapplied out of it because that information is lost. There are infinite solutions to 1 equation and 2 unknowns:
W=Fapplied*Cos(Theta)*displacement
where Fapplied and Theta are unknown.

Is it a matter of me misinterpreting the question? I'm trying but I need more than the book has provided to make sense of this.

Thanks everyone.

2. May 6, 2007

### Office_Shredder

Staff Emeritus
If the parent is supplying 36.666N of force horizontally, we must come to the conclusion the parent is also pulling vertically to an extent, lessening the net force of friction

3. May 6, 2007

### User1247

Ah! that is totally right, so if I allow the parent to be pulling at an angle to the horizontal, Fp would have a vertical component that would cancel out some of Fn and reduce Ff. I'll just pretend that the wagon is a point on the surface of the ground. Thanks, that ought to do it.

4. May 6, 2007

### mbrmbrg

Also, quick note: The problem gives the mass as 50 kg.

5. May 6, 2007

### User1247

Oops. 45 kg has been deeply lodged in my head from my adventures with the last post with the rollerskater going down the hill. Thanks for that mbrmbrg!