# Crate Pushed up an Incline

1. Feb 18, 2010

### easchwen

1. The problem statement, all variables and given/known data
A crate of mass 20 kg is pushed up a ramp by a person, at an angle of 30 degrees. Assume the coefficient of kinetic friction between the crate and the ramp is k=0.2. Calculate the work done by the person in pushing the crate a distance of 29 m as measured along the ramp. Assume that the crate moves with at a constant velocity.

2. Relevant equations
W=mgh
W=F(delta x)cos(theta)=(coefficient of friction)mg(delta x)cos(theta)

3. The attempt at a solution
I found the work due to normal force to be 0, then used the first equation to find the work due to gravity. W=(20kg)(9.8m/s^2)(14.5m)=2842 Next I used the second equation to find the work due to the friction. W=(.2)(20)(9.8)(25.1147)cos(30)=852.59875

I added these two numbers together since the person would have to do work against gravity and the friction and got 3694.59875, but this isn't the right answer. Anyone have help for me? I am thinking my equations might not be right.

2. Feb 18, 2010

### PhanthomJay

For the work done by friction, what's that 25.1147 number???

3. Feb 18, 2010

### easchwen

The 25.1147 is the 29 m times sin(30), or the change in the x direction. The 14.5 is the change in the y direction.

4. Feb 18, 2010

### PhanthomJay

The 14.5 is ok for the y direction when calculating the work done by gravity, but the friction force does not act in the horizontal x direction. What is the formula for work? What is the angle between the friction force and its displacement? What is the value of its displacement?

5. Feb 18, 2010

### easchwen

The formula for work that I found in my notes was W=(coefficient of friction)mg(delta x)cos(theta), so would I just use the 14.5 instead of the 25.1147? I know the displacement of the box is 29 m in the diagonal direction, with an angle of 30 degrees. Am I overthinking this problem? I'm still very confused.

6. Feb 18, 2010

### PhanthomJay

Be careful just looking at your notes without trying to understand WHY work = (umgcostheta)(delta x), which applies only to work being done by forces acting in the same direction of the displacemnt. 'delta x' in this case is the magnitude of the displacement vector between the initial and final position of the crate. How far does the crate move parallel to the incline?

7. Feb 18, 2010

### easchwen

It moves 29 m parallel to the incline... so that's my delta x?

8. Feb 18, 2010

### PhanthomJay

well, yes, but I can see that you are not quite understanding the concept of work. And i'm not sure if you have correctly identified the friction force. So let me ask you:

What is the value of the friction force? Show your equation used in determining that force. Then once you determine that force, what equation would you use to determine the work done by that force?