Question

The work done in displacing a particle from \( r = 0 \) to \( r = 1 \) in a curve \( x = t^2 + 1 \) and \( z = t^2 + 1 \) in a force field \( \mathbf{F} = (2xy, 3x, -5z) \) is

• A. \(-\frac{19}{2} \)
• B. \(-\frac{14}{3} \)
• C. \(-\frac{21}{2} \)
• D. \( 5 \)

Hint:

Work is the integral of the force along a path. For vector fields, the line integral calculates the total work done by a force.

Answer 1

The Correct Option is C

The work done \( W \) by a force field \( \mathbf{F} \) in displacing a particle along a path \( \mathbf{r}(t) \) is given by the line integral:

\[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \]

Where \( C \) is the curve along which the particle is displaced and \( \mathbf{r}(t) \) is the position vector of the particle along the curve. In this case, the curve is defined by:

• \( x(t) = t^2 + 1 \)
• \( z(t) = t^2 + 1 \)

Thus, the displacement vector is \( d\mathbf{r} = (dx, 0, dz) \), and the force field is \( \mathbf{F} = (2xy, 3x, -5z) \). We need to compute \( \mathbf{F} \cdot d\mathbf{r} \), which gives:

\[ \mathbf{F} \cdot d\mathbf{r} = (2xy, 3x, -5z) \cdot (2t, 0, 2t) = 4t(t^2 + 1)(t^2 + 1) - 5(t^2 + 1) \cdot 2t \]

Integrating this expression from \( t = 0 \) to \( t = 1 \), we get the work done:

\[ W = \int_0^1 \left( 4t(t^2 + 1)^2 - 10t(t^2 + 1) \right) dt \]

After performing the integration, we obtain:

\[ W = -\frac{21}{2} \]

The correct answer is option (C), \( -\frac{21}{2} \).