How Much Work is Needed to Pull a Sled on an Icy Road?

[LJ Fiddle]

Homework Statement

By means of a towrope, a college girl pulls a sled loaded with firewood toward her sorority house along a level, icy road. The coefficient of friction between the sled and the road is u_k = 0.12 and the mass of the sled plus its load is 157 kg. The towrope is attached to the front of the sled and makes an angle of 36 degrees with the horizontal. How much work must the girl do with a constant pulling force on the sled to pull it 1.0 km while increasing its velocity from 1.0 m/s to 3.0 m/s?

Homework Equations

Don't know where to start.

The Attempt at a Solution

Friend of mine said: F_SG = (mg)/(sin(theta)*((cos(theta))/u_k))
Using that I got 210 N of force, but with work shouldn't we only be concerned with things in the x-direction?

 

[willem2]

friends don't give friends equations without any explanation,

what are all the forces on the sled? give their horizontal and vertical components.
Since the sled only moves horizontally the vertical forces will do no work, but they
are important for how much friction there is.

 

[LJ Fiddle]

The force the girl exerts on the sled is not given. So that must be found to get the work.

Normal force = mg = 157(9.81) = 1540 N

Friction = u_k * Normal force = 1540 N * 0.12 = 185 N

The angle she is pulling it at is 36 degrees above the horizontal. She pulls it 1.0 km so what is the force over this distance AS SHE INCREASES velocity from 1.0 m/s to 3.0 m/s.

The equation my friend gave me was from a lecture that I missed. He showed my how he got it from $$\Sigma$$F=ma.

I used it and got 210 N. I have no idea if this is right, nor how to proceed from here.

 

[willem2]

The problem is a bit harder than you think. If she pulls upwards, this will reduce the normal force and the friction. Just assume she pulls with a force F. Then you can work out the friction as a function of F and then use $$\Sigma$$ = ma to find what F is.

Once you know F then the horizontal component of F multiplied by the distance she pulls the sled, will give you the work done.

 

[LJ Fiddle]

I see what you mean, by have no idea of going about doing that.

It makes sense because when I use the equation I was shown using angle 0 degrees, I got the frictional force.

 

[devilparticle]

I'm also working on this problem. So far I've figured out the force that the girl pulls the sled because it's just the horizontal component of the force she exerts on the sled. However the fact that her velocity increases across the kilometer has got me thinking that the problem is more complicated that W = FD.

Originally I tried to use the W = delta KE to figure it out but that gives me an answer like 628J, while using W = FD it's more like (185N * cos 36) * 1000m = blah Joules. I'm with the first guy, setting up the problem is where I'm stuck because I can't help thinking the constant acceleration affects the total work.

 

[LJ Fiddle]

I'm also working on this problem. So far I've figured out the force that the girl pulls the sled because it's just the horizontal component of the force she exerts on the sled. However the fact that her velocity increases across the kilometer has got me thinking that the problem is more complicated that W = FD.

Originally I tried to use the W = delta KE to figure it out but that gives me an answer like 628J, while using W = FD it's more like (185N * cos 36) * 1000m = blah Joules. I'm with the first guy, setting up the problem is where I'm stuck because I can't help thinking the constant acceleration affects the total work.

Yup. I'm stuck here too. There's got to be more to this.

 

[willem2]

1. suppose the girl pulls with a force F
2. what is the vertical component of F
3. what must the normal force be to make the forces cancel in the vertical direction?
4. what is the friction on the sled?
5. what is the horizontal component of F?
6. what is the horizontal acceleration of the sled?
7. what must F be to make all the forces in the horizontal direction equal to m*a

the answers of questions 2 through 6 all have the unknown F in them

now the work is (displacement) * (force in the direction of the displacement)

the fact that the sled accelerates is already included in the calculation for the force