A particle is acted upon by a force `vecF=y hati+xhatj` newton. When the particle is moved from (1 m, 1 m) to (9m, 3 m) via straight path, work done by `vecF` is y. What is the value of `x/y?`

Answer

To solve the problem step by step, we will calculate the work done by the force vector \(\vec{F} = y \hat{i} + x \hat{j}\) as the particle moves from point A (1 m, 1 m) to point B (9 m, 3 m).

### Step 1: Identify the force and the displacement
The force acting on the particle is given by:
\[ \vec{F} = y \hat{i} + x \hat{j} \]
The initial position (point A) is \((x_1, y_1) = (1, 1)\) and the final position (point B) is \((x_2, y_2) = (9, 3)\).

### Step 2: Express the differential displacement vector
The differential displacement vector \(d\vec{r}\) can be expressed as:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]

### Step 3: Calculate the work done
The work done \(W\) by the force \(\vec{F}\) when moving from point A to point B is given by the integral:
\[ W = \int_{A}^{B} \vec{F} \cdot d\vec{r} \]
Substituting the expressions for \(\vec{F}\) and \(d\vec{r}\):
\[ W = \int_{x_1}^{x_2} (y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]
This expands to:
\[ W = \int_{x_1}^{x_2} y \, dx + \int_{y_1}^{y_2} x \, dy \]

### Step 4: Set the limits of integration
Here, the limits for \(x\) are from 1 to 9, and for \(y\) from 1 to 3:
\[ W = \int_{1}^{9} y \, dx + \int_{1}^{3} x \, dy \]

### Step 5: Evaluate the integrals
1. For the first integral, since \(y\) is constant (equal to 1 when \(x = 1\)):
\[ W_1 = y \int_{1}^{9} dx = y [x]_{1}^{9} = y(9 - 1) = 8y \]
2. For the second integral, since \(x\) is constant (equal to 1 when \(y = 1\)):
\[ W_2 = x \int_{1}^{3} dy = x [y]_{1}^{3} = x(3 - 1) = 2x \]

### Step 6: Combine the results
Thus, the total work done \(W\) is:
\[ W = 8y + 2x \]
According to the problem, this work done is equal to \(y\):
\[ 8y + 2x = y \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 2x = y - 8y \]
\[ 2x = -7y \]
\[ \frac{x}{y} = -\frac{7}{2} \]

### Final Answer
The value of \(\frac{x}{y}\) is:
\[ \frac{x}{y} = -\frac{7}{2} \]