- #1
KvnBushi
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[SOLVED] diff. paths of a Force in xy-plane
A force acting on a particle in the xy-plane is given by
[tex]\vec{F} = (2yi + x^2j)[/tex] where x and y are in meters.
The particle moves from the origin to a final position having coordinates
[tex]x = 5.00m[/tex] and [tex]y = 5.00m[/tex]
calculate the word done by [tex]F[/tex] on the particle as it moves along
a) [tex]OAC[/tex] ------------- solved
b) [tex]OBC[/tex] ------------- solved
c) [tex]OC[/tex] ---------------- need help
Here is the xy-plane graphed with each point ( ignore the white ) :
B|......C ( 5.00m , 5.00m )
..|
..|
..|______________
O.....A ( 5.00m , 0.00m )
[tex]W = |F| \Delta r cos\theta[/tex]
[tex]W_{x} = \int_a^b F_{x} dx[/tex]
[tex]W = \vec{F}_{x} \cdot \Delta \vec{r}[/tex]
a)
[tex]W_O_A = <2y, x^2> \cdot <5,0> = 10y = 10(0) = 0[/tex]
[tex]W_A_C = <2y, x^2> \cdot <0,5> = 5x^2 = 5(5)^2 = 125[/tex]
[tex]W_O_A_C = 0 + 125 = 125 J[/tex]
b)
[tex]W_O_B = <2y, x^2> \cdot <0,5> = 5x^2 = 5(0)^2 = 0[/tex]
[tex]W_B_C = <2y, x^2> \cdot <5,0> = 10y = 10(5) = 50[/tex]
[tex]W_O_B_C = 0 + 50 = 50J[/tex]
c)
I attempted this using 3 different equations, all of which gave me different answers and none of which were the correct answer.
Correct answer: 66.7 J
1. [tex]W_O_C = <2y, x^2> \cdot <5,5> = 10y + 5x^2 = 10(5) + 5(5)^2 = 50 + 125 = 175J[/tex]
2. [tex]W_O_C = |F| \Delta r cos \theta = \sqrt{4y^2 + x^4}(7.04)(cos0) = 7.07 \sqrt{4(5)^2 + (5)^4} = 7.07 sqrt{725} = 190.34 J[/tex]
3. [tex]W_O_C = \int_0^5 x dx = \frac{1}{2} x^2 |_0^5 = \frac{1}{2} (125) = 12.5 J [/tex]
I can find logical faults in each of my approaches, but I do not know what else to do. I would highly appreciate hints/answers to the right approach, as well as why my above approaches are inappropriate.
One way I would try to solve this, using Calc 3 (which should not necessary for this level physics) would be something like this:
[tex]\int_0^5 \int_0^5 F_x_y dy dx = \int_0^5 \int_0^5 \sqrt{4y^2i + x^4j } dy dx[/tex]
But at first look it seems pretty difficult.
Thank you very much,
Kevin.
Homework Statement
A force acting on a particle in the xy-plane is given by
[tex]\vec{F} = (2yi + x^2j)[/tex] where x and y are in meters.
The particle moves from the origin to a final position having coordinates
[tex]x = 5.00m[/tex] and [tex]y = 5.00m[/tex]
calculate the word done by [tex]F[/tex] on the particle as it moves along
a) [tex]OAC[/tex] ------------- solved
b) [tex]OBC[/tex] ------------- solved
c) [tex]OC[/tex] ---------------- need help
Here is the xy-plane graphed with each point ( ignore the white ) :
B|......C ( 5.00m , 5.00m )
..|
..|
..|______________
O.....A ( 5.00m , 0.00m )
Homework Equations
[tex]W = |F| \Delta r cos\theta[/tex]
[tex]W_{x} = \int_a^b F_{x} dx[/tex]
[tex]W = \vec{F}_{x} \cdot \Delta \vec{r}[/tex]
The Attempt at a Solution
a)
[tex]W_O_A = <2y, x^2> \cdot <5,0> = 10y = 10(0) = 0[/tex]
[tex]W_A_C = <2y, x^2> \cdot <0,5> = 5x^2 = 5(5)^2 = 125[/tex]
[tex]W_O_A_C = 0 + 125 = 125 J[/tex]
b)
[tex]W_O_B = <2y, x^2> \cdot <0,5> = 5x^2 = 5(0)^2 = 0[/tex]
[tex]W_B_C = <2y, x^2> \cdot <5,0> = 10y = 10(5) = 50[/tex]
[tex]W_O_B_C = 0 + 50 = 50J[/tex]
c)
I attempted this using 3 different equations, all of which gave me different answers and none of which were the correct answer.
Correct answer: 66.7 J
1. [tex]W_O_C = <2y, x^2> \cdot <5,5> = 10y + 5x^2 = 10(5) + 5(5)^2 = 50 + 125 = 175J[/tex]
2. [tex]W_O_C = |F| \Delta r cos \theta = \sqrt{4y^2 + x^4}(7.04)(cos0) = 7.07 \sqrt{4(5)^2 + (5)^4} = 7.07 sqrt{725} = 190.34 J[/tex]
3. [tex]W_O_C = \int_0^5 x dx = \frac{1}{2} x^2 |_0^5 = \frac{1}{2} (125) = 12.5 J [/tex]
I can find logical faults in each of my approaches, but I do not know what else to do. I would highly appreciate hints/answers to the right approach, as well as why my above approaches are inappropriate.
One way I would try to solve this, using Calc 3 (which should not necessary for this level physics) would be something like this:
[tex]\int_0^5 \int_0^5 F_x_y dy dx = \int_0^5 \int_0^5 \sqrt{4y^2i + x^4j } dy dx[/tex]
But at first look it seems pretty difficult.
Thank you very much,
Kevin.