Find the work given in photo F = (3xi^+4j^) x is in meters particle moves from a position (2m,3m) to (3m,0m)

To find the work done by the force field \( \mathbf{F} = (3x \hat{i} + 4 \hat{j}) \) as a particle moves from the position \( (2 \, \text{m}, 3 \, \text{m}) \) to \( (3 \, \text{m}, 0 \, \text{m}) \), we can use the concept of line integrals. 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 Field

The force field is defined as:

F = (3x) i + (4) j

Here, \( x \) is the x-coordinate of the particle's position, and \( \hat{i} \) and \( \hat{j} \) are the unit vectors in the x and y directions, respectively.

Path of the Particle

The particle moves from the point \( (2, 3) \) to \( (3, 0) \). We can describe this path parametrically. A simple way to do this is to use a linear interpolation between the two points:

• Let \( x(t) = 2 + t \) where \( t \) varies from \( 0 \) to \( 1 \).
• For the y-coordinate, we can interpolate as \( y(t) = 3 - 3t \).

Thus, the parametric equations for the path are:

x(t) = 2 + t

y(t) = 3 - 3t

where \( t \) ranges from \( 0 \) to \( 1 \).

Calculating the Work Done

The work done by the force along the path is given by the line integral:

W = ∫ F \cdot dr

Here, \( dr \) is the differential displacement vector along the path. We can express \( dr \) in terms of \( dt \):

dr = (dx, dy) = (dx/dt \, dt, dy/dt \, dt)

Calculating the derivatives:

• From \( x(t) = 2 + t \), we have \( dx/dt = 1 \).
• From \( y(t) = 3 - 3t \), we have \( dy/dt = -3 \).

Thus, we can express \( dr \) as:

dr = (1 \, dt, -3 \, dt) = (dt, -3 \, dt)

Substituting into the Integral

Now, we substitute \( x(t) \) into the force field:

F(t) = (3(2 + t) \hat{i} + 4 \hat{j}) = (6 + 3t) \hat{i} + 4 \hat{j}

Now we can compute the dot product \( F \cdot dr \):

F \cdot dr = ((6 + 3t) \hat{i} + 4 \hat{j}) \cdot (dt, -3 \, dt) = (6 + 3t) \cdot 1 \, dt + 4 \cdot (-3) \, dt

This simplifies to:

F \cdot dr = (6 + 3t - 12) \, dt = (3t - 6) \, dt

Evaluating the Integral

The work done is then:

W = ∫ (3t - 6) \, dt

We evaluate this integral from \( t = 0 \) to \( t = 1 \):

W = ∫ (3t - 6) \, dt = \left[ \frac{3}{2}t^2 - 6t \right]_{0}^{1}

Calculating the limits:

• At \( t = 1 \): \( \frac{3}{2}(1)^2 - 6(1) = \frac{3}{2} - 6 = -\frac{9}{2} \).
• At \( t = 0 \): \( \frac{3}{2}(0)^2 - 6(0) = 0 \).

Thus, the work done is:

W = -\frac{9}{2} - 0 = -\frac{9}{2} \, \text{Joules}

Final Result

The work done by the force field as the particle moves from \( (2 \, \text{m}, 3 \, \text{m}) \) to \( (3 \, \text{m}, 0 \, \text{m}) \) is \( -\frac{9}{2} \, \text{J} \) or \( -4.5 \, \text{J} \). This negative value indicates that the force is doing work against the motion of the particle.