A particle is traveling in a straight line at a constant speed of 27.3 m/s. Suddenly, a constant force of 14.3 N acts on it, bringing it to a stop in a distance of 56.4 m.

To find the acceleration of the particle, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

where:
- vf is the final velocity (0 m/s, since the particle comes to a stop),
- vi is the initial velocity (27.3 m/s),
- a is the acceleration of the particle,
- d is the distance traveled (56.4 m).

Rearranging the equation, we get:

a = (vf^2 - vi^2) / (2d)

Plugging in the values, we have:

a = (0^2 - 27.3^2) / (2 * 56.4) = -746.55 / 112.8

a ≈ -6.625 m/s^2

The negative sign indicates that the deceleration (acceleration in the opposite direction of motion) is acting on the particle.

To find the mass of the particle, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, we need to find the mass of the particle. We know the force acting on the particle is 14.3 N, and the final velocity is 0 m/s (since the particle comes to a stop). However, we don't have the acceleration.

To find the acceleration, we can use the kinematic equation that relates distance, initial velocity, final velocity, and acceleration:

v^2 = u^2 + 2as

Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the distance

Plugging in the given values:
0^2 = 27.3^2 + 2 * a * 56.4

Simplifying:
0 = 745.29 + 112.8a

Rearranging the equation:
112.8a = -745.29

Dividing both sides by 112.8:
a = -6.6 m/s^2

Now that we have the acceleration, we can substitute it back into Newton's second law of motion:

14.3 N = m * (-6.6 m/s^2)

Solving for mass:
m = 14.3 N / (-6.6 m/s^2)
m ≈ -2.17 kg

The mass of the particle is approximately 2.17 kg.