If F =(xi +yj) acts on a particle.then find the work done by force in displacing the particle from (0,0) to (2,2)

To find the work done by the force \( F = (xi + yj) \) when displacing a particle from the point \( (0,0) \) to \( (2,2) \), we can use the concept of line integrals in vector calculus. The work done by a force along a path is given by the integral of the force along that path. Let's break this down step by step.

Understanding the Force and Path

The force vector \( F \) can be expressed in terms of its components as:

• \( F_x = x \) (the component in the x-direction)
• \( F_y = y \) (the component in the y-direction)

We are moving the particle from the initial position \( (0,0) \) to the final position \( (2,2) \). The path taken can be described parametrically. A simple way to represent this linear path is to use the parameter \( t \), where \( t \) varies from 0 to 1:

• \( x(t) = 2t \)
• \( y(t) = 2t \)

Calculating the Differential Displacement

The differential displacement vector \( dr \) along the path can be expressed as:

• \( dr = (dx, dy) = (2dt, 2dt) \)

Setting Up the Work Integral

The work done \( W \) by the force along the path is given by the line integral:

\( W = \int_C F \cdot dr \)

Substituting the expressions for \( F \) and \( dr \), we have:

\( W = \int_0^1 F(x(t), y(t)) \cdot (dx, dy) \)

Now substituting \( F \) and \( dr \):

\( F(x(t), y(t)) = (2t i + 2t j) \)

Thus, the dot product \( F \cdot dr \) becomes:

\( F \cdot dr = (2t i + 2t j) \cdot (2dt i + 2dt j) = (2t)(2dt) + (2t)(2dt) = 4t dt + 4t dt = 8t dt \)

Evaluating the Integral

Now we can evaluate the integral:

\( W = \int_0^1 8t \, dt \)

Calculating this integral:

\( W = 8 \left[ \frac{t^2}{2} \right]_0^1 = 8 \left[ \frac{1^2}{2} - \frac{0^2}{2} \right] = 8 \cdot \frac{1}{2} = 4 \)

Final Result

The work done by the force \( F \) in displacing the particle from \( (0,0) \) to \( (2,2) \) is \( 4 \) units of work. This result illustrates how we can use vector calculus to compute work done by a force along a specified path, integrating the force along the trajectory of the particle.