# Homework Help: Work done by a changing force

1. Jan 9, 2014

### KiNGGeexD

I am stuck with the above question, my answer to a is 40J but the model answer is 30J

Really frustrating as the book I am ready only goes over straight line graphs

2. Jan 9, 2014

### NasuSama

Remember that the work done for this type of problem is the area between the $x$-axis and the given curve. You got that value via adding up the areas between $x = -2$ and $x = 4$. The first part asks you to determine the total work done. You also need to consider the "negative" work done between $x = -4$ and $x = -2$. It also takes in account for this problem.

Try again to see if you can get $30$ J instead of $40$ J.

3. Jan 9, 2014

### KiNGGeexD

So for the first curve can I use 1/2B*h?

Because I done that for the two slopes and b*h for the square part?

4. Jan 9, 2014

### NasuSama

Yes, you need to use the area of the triangle formula for the region under the $x$-axis to obtain the "negative" work done.

Since we want to compute the total work done between $x = -4$ and $x = 4$, we combine the works altogether to obtain the result. Don't forget this part.

5. Jan 9, 2014

### KiNGGeexD

So I get 4 work values?

6. Jan 9, 2014

### KiNGGeexD

So from left to right I get

20J
-10J
20J
10J

So that gives me 40J?

7. Jan 9, 2014

### NasuSama

I didn't get those values. Let's try again.

Remember that for the region under the $x$-axis, regardless of the values of $x$, you get the negative work done, using the area of the triangle:

$A = -\dfrac{1}{2}|b||h|$

where the brackets indicate the absolute value of any number.

Then, for the region above the $x$-axis, (also regardless of the values of $x$) you use the area of the trapezoid, which states

$A = \dfrac{1}{2}|h||b_1 + b_2|$

where $b_1$ and $b_2$ are arbitrary bases of the trapezoid and $h$ is the height of the trapezoid.

You can also compute the area of the bigger region by triangle and square area formulas as you attempted. That is: determine the area of two triangles and the area of the square.

So in summary, you should get the following equation (ignoring the signs of $x$ values)

$\text{Total work done} = \text{Area between x = -4 and x = -2} + \text{Area between x = -2 and x = 4}$

where $\text{Area between x = -4 and x = -2}$ is negative while other area is positive.

Notes: Be careful about how you compute the first value. The shape is a triangle, not a square! Also be very careful about the signs of the first two values. If a region occurs under $x$-axis, then the area is negative; otherwise, it's positive.

Last edited: Jan 9, 2014
8. Jan 9, 2014

### KiNGGeexD

Ah ok I think I got it:)