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

[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?

 

[Pyrrhus]

By the definition of Work

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

 

[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 isn't a change in work

 

[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.

 

[UrbanXrisis]

that doesn't 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?

 

[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:

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

 

[UrbanXrisis]

I don't understand how the force can be different. Do you have an example?

 

[Pyrrhus]

that doesn't 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?

What do you mean not getting different answers?

Probably your limits are wrong.

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.

 

[UrbanXrisis]

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

 

[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:

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

 

[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
$$\int_{0}^{5} 2y dx =y^2 = 25$$

 

[Pyrrhus]

It was disregarded because of

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

and read above

 

[UrbanXrisis]

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

Is there another way to approach this without integrals?

 

[Pyrrhus]

Not that i know of, urban.