# What is the force work done by force F along OAC, OBC, OC

1. Dec 13, 2004

### UrbanXrisis

B ------C (5,5)
..|...../|
..|.../..|
O|/___|A

A force acting on a particle moving in the xy plane is given by F=(2yi + x^2j) N, where x and y are in meters. THe particle moves from the origin to a position having coordinates x=5.00m and y=5.00m.

What is the force work done by force F along OAC, OBC, OC.

The answer books shows different answer. Woudln't the work just be mgh for all of them? Obvioulsy not, how would I solve this problem?

Last edited: Dec 13, 2004
2. Dec 13, 2004

### Pyrrhus

By the definition of Work

$$W = \int \vec{F } \cdot d \vec{r}$$

3. Dec 13, 2004

### UrbanXrisis

the integral of F=(2yi + x^2j) is

F=y^2+.3x^3

so W=(y^2+.3x^3)

however, (x,y) is always 5,5, there isnt a change in work

4. Dec 13, 2004

### Pyrrhus

No, that's not true, for the first case you must evaluate from O to B, then from B to C and add both works for the OBC path.

5. Dec 13, 2004

### UrbanXrisis

that doesnt seem to get me any new answers.

O-->(0,5)B-->(5,0)C is W=(0^2+.3(5)^3)+(5^2+.3(0)^3)

which equals (5^2+.3(5)^3)=66.6

how is it possible to get different numbers?

6. Dec 13, 2004

### Pyrrhus

Because that force is not conservative, it depends on the trajectory.

This means that the work done by the particle on a closed circuit or on a path from 1 to 2 and then 2 to 1 depends on the trajectory is not equal to 0, thus the force is not conservative.

The following condition is not met:

$$\oint \vec{F} \cdot d \vec{s} = 0$$

7. Dec 13, 2004

### UrbanXrisis

I dont understand how the force can be different. Do you have an example?

8. Dec 13, 2004

### Pyrrhus

What do you mean not getting different answers???

For example for the OBC Path

You first find the work from OB which means to Integrate from 0 to 5 with a differential dy

$$W_{ob} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) \cdot dy \hat_{j} = \int_{0}^{5} x^2 dy = 0$$

because x = 0 along this path

Now BC

$$W_{bc} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) \cdot dx \hat_{i} = \int_{0}^{5} 2y dx = 50$$

because y = 5

The work along OBC is 50 Joules.

Last edited: Dec 13, 2004
9. Dec 13, 2004

### UrbanXrisis

why is OB=0? there is a change in height.

10. Dec 13, 2004

### Pyrrhus

Because x = 0 and is considered a constant in the integral, what is considered a variable is what the differential tell us.

$$W_{ob} = x^2 \int_{0}^{5} dy$$

Also Remember:

$$\hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = \hat{j} \cdot \hat{k} = 0$$

Last edited: Dec 13, 2004
11. Dec 13, 2004

### UrbanXrisis

for:
$$W_{ob} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) dy \hat_{j} = \int_{0}^{5} x^2 dy = 0$$

why was the y value disregarded?

for:
$$W_{bc} = \int_{0}^{5} (2y \hat{i} + x^2 \hat{j}) dx \hat_{i} = \int_{0}^{5} 2y dx = 50$$

how does:
$$\int_{0}^{5} 2y dx = 50$$
when
$$\int_{0}^{5} 2y dx =y^2 = 25$$

12. Dec 13, 2004

### Pyrrhus

It was disregarded because of

$$\hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = \hat{j} \cdot \hat{k} = 0$$

13. Dec 13, 2004

### UrbanXrisis

I'm sorry but I have no idea what that means

Is there another way to approach this without integrals?

14. Dec 13, 2004

### Pyrrhus

Not that i know of, urban.