Angular Momentum of the Particle

The angular momentum (L) of a particle is given by the cross product of its position vector (r) and its linear momentum (p). Mathematically, it can be expressed as:

L = r × p

Here, r is the position vector and p is the linear momentum vector. In this case, the particle is moving on a straight line, so its position vector can be represented as r = xi, where x is the position of the particle on the x-axis.

Given that the angular momentum of the particle about the origin is L = 3t i, we can equate it to the cross product of r and p:

L = r × p = 3t i

Since the particle is moving on a straight line, its linear momentum vector will be in the same direction as the position vector, i.e., p = p xi.

Cross Product Calculation

To determine the force acting on the particle at t = 2 sec, we need to find the cross product of r and p. Since both r and p are in the x-direction, the cross product simplifies to:

L = r × p = (xi) × (pxi) = (x * p) (i × i)

As the cross product of any vector with itself is zero, we have:

L = (x * p) (i × i) = 0

Therefore, the angular momentum of the particle is zero at all times. This implies that there is no torque acting on the particle, and hence, no net force acting on it.

Conclusion

In conclusion, the force acting on the particle at t = 2 sec is zero. This is because the angular momentum of the particle about the origin is zero, indicating that there is no torque acting on the particle. As a result, there is no net force acting on the particle.

Note: This answer is provided based on the given information. If any additional details or constraints are provided, the solution may vary accordingly.