# Child dragging box (friction, net force and work)

1. Mar 9, 2010

### Jim01

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

A child drags a 20kg box across a lawn for 10m anfd along a sidewalk for 30m; the coefficient of friction is 0.25 for the first part of the trip and 0.55 for the second part. If the child always pulls horizontally, how much work does the child do on the box?

2. Relevant equations

W = Fnet,x$$\Delta$$x

3. The attempt at a solution

The only way that I know how to find a net force is to find the summation of F for part one and the summation of F for part 2. While the problem dioesn't say so, I am going to assume there is constant velocity.

$$\Sigma$$Fx1=0

Wx1=0
Px1= ?
Ff1= -$$\mu$$k1mg

Px1 - $$\mu$$k1mg = 0

$$\Sigma$$Fy1=0

Wy1= -mg
N1= mg

mg - mg = 0

$$\Sigma$$Fx2=0

Wx2=0
Px2= ?
Ff2= -$$\mu$$k2mg

Px2 - $$\mu$$k2mg = 0

$$\Sigma$$Fy2=0

Wy2= -mg
N2= mg

mg-mg=0

So here is where I get lost. I can find Px1 (49.05N) and Px2 (107.91N). Do I just add them together and plug them into the F portion of the W = Fnet,x$$\Delta$$x formula, using 40m as my ]$$\Delta$$x?

2. Mar 9, 2010

### Fightfish

I think you are making things unnecessarily complicated here...the working definitely can be simplified somewhat.
In any case, you are right in obtaining the (pulling) force along the lawn as 49.05N and that along the sidewalk as 107.91N. However, you cannot add them together and plug them into the equation for work! That does not make any sense. Rather, you have to evaluate the work done on the box for each part (lawn and sidewalk) separately, and add the work together to obtain the total work done by the child on the box.

3. Mar 9, 2010

### Jim01

I sometimes have problems with my summation of forces and find that if I seperate the x and y forces and label everything it helps me. I agree that it is tedious, but I tend to not "see" things and leave them out if I do not. I am hoping that as I continue on and get more comfortable with the process I will no longer need to do it.

Once again, thank you for your help. I thought that the formula W = Fnet,x$$\Delta$$x meant that I was to add all of the forces and apply them to the formula.

Instead I see now that I will have a W1 = F1,x $$\Delta$$x1 and a W2 = F2,x $$\Delta$$x2 and that Wt = W1 + W2

4. Mar 9, 2010

### PhanthomJay

Good observation. You really couldn't solve the problem without more info if the box was accelerating, but nonetheless, your assumption shows keen insight. The problem should state such.