Constant Force Particle Motion Analysis

The question asks for the proportionality of the velocity attained by a stationary particle of mass '$m$' when a constant force '$F$' is applied over a certain displacement '$s$'. We can solve this using basic physics principles.

Deriving Velocity Proportionality

• Newton's Second Law: The force '$F$' applied to the particle causes acceleration '$a$'. Using Newton's second law, $F = ma$.
• Calculate Acceleration: Since the force '$F$' and mass '$m$' are constant, the acceleration '$a$' is also constant. We can express acceleration as:

  $a = \frac{F}{m}$

• Kinematic Equation: We use a standard kinematic equation that relates final velocity '$v$', initial velocity '$u$', acceleration '$a$', and displacement '$s$'. The relevant equation is:

  $v^2 = u^2 + 2as$

• Apply Initial Conditions: The particle starts from rest, so the initial velocity '$u = 0$'. Substituting this into the equation gives:

  $v^2 = 0^2 + 2as$

  $v^2 = 2as$

• Substitute Acceleration: Now, substitute the expression for acceleration ($a = F/m$) into the equation:

  $v^2 = 2 \left( \frac{F}{m} \right) s$

• Find Velocity: Taking the square root to find the velocity '$v$':

  $v = \sqrt{\frac{2Fs}{m}}$

• Determine Proportionality: The question asks what the velocity '$v$' is proportional to. In the equation $v = \sqrt{\frac{2Fs}{m}}$, the force '$F$', the displacement '$s$', and the constant '2' are considered fixed for this specific scenario. Therefore, the velocity '$v$' is inversely proportional to the square root of the mass '$m$'.

  $v \propto \sqrt{\frac{1}{m}}$

  $v \propto \frac{1}{\sqrt{m}}$

Conclusion

The velocity attained by the particle is proportional to $\frac{1}{\sqrt{m}}$.