A force `vecF=k[yhati+xhatj]` where k is a positive constant acts on a particle moving in x-y plane starting from the point (3, 5), the particle is taken along a straight line to (5, 7). The work done by the force is :

Options

A. zero
B. 35 K
C. 20 K
D. 15 K

Answer

To find the work done by the force \(\vec{F} = k(y \hat{i} + x \hat{j})\) on a particle moving from point \(A(3, 5)\) to point \(B(5, 7)\), we can follow these steps:

### Step 1: Define the Force
The force acting on the particle is given by:
\[ \vec{F} = k(y \hat{i} + x \hat{j}) \]
where \(k\) is a positive constant, \(y\) is the y-coordinate, and \(x\) is the x-coordinate of the particle.

### Step 2: Parameterize the Path
The particle moves from point \(A(3, 5)\) to point \(B(5, 7)\). We can parameterize the path as follows:
Let \(x(t) = 3 + 2t\) and \(y(t) = 5 + 2t\) where \(t\) varies from \(0\) to \(1\). This gives us a straight line path from \(A\) to \(B\).

### Step 3: Calculate the Differential Displacement
The differential displacement vector \(d\vec{r}\) is given by:
\[ d\vec{r} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} = (2 \hat{i} + 2 \hat{j}) dt \]

### Step 4: Substitute the Parameterization into the Force
Substituting \(x(t)\) and \(y(t)\) into the force:
\[ \vec{F}(t) = k((5 + 2t) \hat{i} + (3 + 2t) \hat{j}) \]

### Step 5: Calculate the Work Done
The work done \(W\) by the force along the path from \(A\) to \(B\) is given by the integral:
\[ W = \int_A^B \vec{F} \cdot d\vec{r} \]
Substituting the expressions we have:
\[ W = \int_0^1 \vec{F}(t) \cdot (2 \hat{i} + 2 \hat{j}) dt \]
Calculating the dot product:
\[ \vec{F}(t) \cdot d\vec{r} = k((5 + 2t) \cdot 2 + (3 + 2t) \cdot 2) dt = k(2(5 + 2t) + 2(3 + 2t)) dt \]
\[ = k(10 + 4t + 6 + 4t) dt = k(16 + 8t) dt \]
Now, integrate:
\[ W = k \int_0^1 (16 + 8t) dt = k \left[ 16t + 4t^2 \right]_0^1 = k(16 + 4) = 20k \]

### Final Answer
Thus, the work done by the force is:
\[ W = 20k \]