Work Done by Force from 0 to 2 m - 20.4J

[maniacp08]

Homework Statement

a) What is the work done by the force from x = 0 to x = 2 m?

Homework Equations

Work = Integral of Force Dx

The Attempt at a Solution

I was told since the equation for Force is not given, my best bet is to count the boxes under the curve from x=0 to x=2. I counted a total approx. of 20.4 boxes = 20.4J am I doing something wrong?
Thanks for helping.

 

[Hootenanny]

Homework Statement

a) What is the work done by the force from x = 0 to x = 2 m?

Homework Equations

Work = Integral of Force Dx

The Attempt at a Solution

I was told since the equation for Force is not given, my best bet is to count the boxes under the curve from x=0 to x=2. I counted a total approx. of 20.4 boxes = 20.4J am I doing something wrong?
Thanks for helping.

Is each box equivalent 1J?

 

[maniacp08]

Y axis is measured in Newtons
X axis is measured in meters

isn't 1 full box, is a Newton*Meters = 1 J?

 

[Hootenanny]

Y axis is measured in Newtons
X axis is measured in meters

isn't 1 full box, is a Newton*Meters = 1 J?

Look at the scale on the graph.

How many increments per meter? How many increments per Newton?

 

[maniacp08]

Oh, is .25m increment for x-axis and .5Newton increment for y axis.

If I counted to be approx. 20.4 boxes
How would I calculate the work?

Or I should do each box?
say first .25m is .5n * .25m and
.50m is .50m * 1n?

 

[Hootenanny]

First work out how much each box is worth (i.e. 0.25*0.5 Joules) and then multiply this by the total number of boxes.

 

[maniacp08]

Thanks Hootenanny!
Time to start cracking on those graph problems.

 

[Hootenanny]

Thanks Hootenanny!
Time to start cracking on those graph problems.

No problem