A force acting on a particle moving in the xy plane is given by Fx = (2yi +x^2j)N, where x and y are in meters. The particle moves from the origin to a final position having coordinates x=5.00m and y=5.00m, as in Fig. Calculate the work done by F along (a) OAC, (b) OBC, (c) OC.(d) Is F conservative or nonconservative? Explain.
yes
To calculate the work done by the force along different paths, we need to use the formula:
\[ W = \int \vec{F} \cdot d\vec{r} \]
where:
- \(\vec{F}\) is the force vector,
- \(d\vec{r}\) is the displacement vector.
Given that \( \vec{F_x} = (2y\hat{i} + x^2\hat{j}) \) N, where x and y are in meters, we can rewrite it as:
\[ \vec{F} = (2y\hat{i} + x^2\hat{j}) \]
Now, let's calculate the work done along each path:
(a) OAC:
In this case, the position vector \(\vec{r}\) varies only along the x-axis while keeping y constant at 0. So, \(d\vec{r} = dx\hat{i}\). Therefore,
\[ W_{OAC} = \int_{O}^{A} \vec{F} \cdot d\vec{r} \]
\[ W_{OAC} = \int_{0}^{5} (2(0)\hat{i} + x^2\hat{j}) \cdot (dx\hat{i}) \]
\[ W_{OAC} = \int_{0}^{5} (x^2) \cdot dx \]
\[ W_{OAC} = \left[ \frac{x^3}{3} \right]_{0}^{5} \]
\[ W_{OAC} = \frac{5^3}{3} - \frac{0^3}{3} \]
\[ W_{OAC} = \frac{125}{3} \]
(b) OBC:
In this case, the position vector \(\vec{r}\) varies only along the y-axis while keeping x constant at 5. So, \(d\vec{r} = dy\hat{j}\). Therefore,
\[ W_{OBC} = \int_{O}^{B} \vec{F} \cdot d\vec{r} \]
\[ W_{OBC} = \int_{0}^{5} (2y\hat{i} + (5^2)\hat{j}) \cdot (dy\hat{j}) \]
\[ W_{OBC} = \int_{0}^{5} (25) \cdot dy \]
\[ W_{OBC} = \left[ 25y \right]_{0}^{5} \]
\[ W_{OBC} = 25(5) - 25(0) \]
\[ W_{OBC} = 125 \]
(c) OC:
In this case, the position vector \(\vec{r}\) varies in both x and y directions. So, \(d\vec{r} = dx\hat{i} + dy\hat{j}\). Therefore,
\[ W_{OC} = \int_{O}^{C} \vec{F} \cdot d\vec{r} \]
\[ W_{OC} = \int_{0}^{5} (2y\hat{i} + x^2\hat{j}) \cdot (dx\hat{i} + dy\hat{j}) \]
⇒ Since the above equation is a dot product, we can break it down into the sum of two integrals:
\[ W_{OC} = \int_{0}^{5} (2y\hat{i}) \cdot (dx\hat{i}) + \int_{0}^{5} (x^2\hat{j}) \cdot (dy\hat{j}) \]
The first integral evaluates to 0 because there is no component of \(dx\hat{i}\) along the y-axis.
The second integral evaluates to:
\[ \int_{0}^{5} (x^2\hat{j}) \cdot (dy\hat{j}) = \int_{0}^{5} (x^2) \cdot dy \]
\[ = \left[ x^2y \right]_{0}^{5} \]
\[ = 5^2(5) - 0^2(0) \]
\[ = 125 \]
Therefore,
\[ W_{OC} = 125 \]
(d) To determine if the force is conservative or nonconservative, we need to check if the work done along a closed path is zero. In this case, the only closed paths are OACO and OBCO:
For OACO, we have already evaluated the work done to be \( \frac{125}{3} \).
For OBCO, the force is constant along the path and independent of position. Therefore, work done will also be zero.
Since the work done along OACO is not zero, the force is nonconservative.
To calculate the work done by a force along a particular path, we need to integrate the dot product of the force and the displacement along that path.
In this case, we are given the force Fx = (2yi + x^2j)N, where x and y are in meters. The particle moves from the origin to a final position with coordinates x=5.00m and y=5.00m.
(a) To calculate the work done along OAC, we need to consider the path from the origin (O) to point A and then to point C.
The displacement vector along OAC can be calculated by subtracting the position vector of O from the position vector of C:
dA = (5.00i + 5.00j)m - (0i + 0j)m = (5.00i + 5.00j)m
Now, we can calculate the work done by integrating the dot product of the force and the displacement vector along OAC:
W_OAC = ∫ (Fx · dA)
Substituting the given force equation:
W_OAC = ∫ ((2y)i + (x^2)j) · (5.00i + 5.00j) dx
Expanding the dot product:
W_OAC = ∫ (10y + 5x^2) dx
The limits of integration for this path would depend on the specific boundaries in the figure you mentioned.
(b) To calculate the work done along OBC, we need to consider the path from point O to point B and then to point C.
The displacement vector along OBC can be calculated by subtracting the position vector of O from the position vector of C:
dB = (5.00i + 5.00j)m - (0i + 0j)m = (5.00i + 5.00j)m
Now, we can calculate the work done by integrating the dot product of the force and the displacement vector along OBC:
W_OBC = ∫ (Fx · dB)
Substituting the given force equation:
W_OBC = ∫ ((2y)i + (x^2)j) · (5.00i + 5.00j) dx
Expanding the dot product:
W_OBC = ∫ (10y + 5x^2) dx
The limits of integration for this path would also depend on the specific boundaries in the figure you mentioned.
(c) To calculate the work done along OC, we need to consider the direct path from point O to point C.
The displacement vector along OC can be calculated by subtracting the position vector of O from the position vector of C:
dC = (5.00i + 5.00j)m - (0i + 0j)m = (5.00i + 5.00j)m
Now, we can calculate the work done by integrating the dot product of the force and the displacement vector along OC:
W_OC = ∫ (Fx · dC)
Substituting the given force equation:
W_OC = ∫ ((2y)i + (x^2)j) · (5.00i + 5.00j) dx
Expanding the dot product:
W_OC = ∫ (10y + 5x^2) dx
The limits of integration for this path would also depend on the specific boundaries in the figure you mentioned.
(d) To determine if the force F is conservative or nonconservative, we need to evaluate if the work done by the force depends on the path taken or only on the initial and final positions.
If the work done by F along any closed path is zero, then F is conservative. If the work done depends on the path, then F is nonconservative.
To assess this, we would need to compare the values of the calculated work done for each of the paths (OAC, OBC, OC). If the work done along each path is the same, regardless of the path taken, then the force F is conservative. If the work done differs for each path, the force F is nonconservative.
Please note that the specific boundaries and constraints of the figure mentioned are required to provide a more precise calculation and assessment of the work done and the nature of the force.
If you tried to copy and paste, that does not work here. Therefore I do not know your paths.
If the force is conservative the curl of the force vector is zero and it can be describedas the gradient of a potential. assuming you mean F = 2y i +x^2 j
i j k
d/dx d/dy d/dz
2 y x^2 0
[d/dx(x^2) - d/dy(2y)] k = (2x-2)k
see http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=85afa8b72bd564fc6c84dde9f0393305&title=MathsPro101%20-%20%20Curl%20and%20Divergence%20of%20Vector%20Fields&theme=gray&i0=2y&i1=x%5E2&i2=0&i3=curl&podSelect=&includepodid=VectorAnalysisResult
so
not a conservative field and the work done will depend on the path taken