Physics Analysis: Constant Power Motion

This question asks for the relationship between a block's speed (\(v\)) and time (\(t\)) when pulled by a machine exerting constant power (\(P\)) on a smooth horizontal surface.

Governing Physics Principles

• Power: The rate at which work is done. For a force \(F\) acting on an object moving at velocity \(v\), power is \(P = F \cdot v\).
• Constant Power: The problem states \(P\) is constant. Thus, \(F \cdot v = P\), which implies \(F = \frac{P}{v}\).
• Newton's Second Law: The net force on the object equals mass (\(m\)) times acceleration (\(a\)), \(F_{net} = ma\). On a smooth surface, \(F_{net} = F\).
• Acceleration: Acceleration is the rate of change of velocity, \(a = \frac{dv}{dt}\).

Derivation Steps

1. Combine the force equations: \(ma = \frac{P}{v}\).
2. Substitute \(a = \frac{dv}{dt}\): \( m \frac{dv}{dt} = \frac{P}{v} \)
3. Separate variables to solve the differential equation. Assume the block starts from rest (\(v=0\) at \(t=0\)). \( m v \, dv = P \, dt \)
4. Integrate both sides: \( \int_{0}^{v} m v' \, dv' = \int_{0}^{t} P \, dt' \) \( \left[ \frac{1}{2} m v'^2 \right]_{0}^{v} = \left[ P t' \right]_{0}^{t} \) \( \frac{1}{2} m v^2 = P t \)
5. Solve for \(v\) in terms of \(t\): \( v^2 = \frac{2P}{m} t \) \( v = \sqrt{\frac{2P}{m}} \sqrt{t} \)

The derived relationship shows that speed \(v\) is proportional to the square root of time, \(v \propto \sqrt{t}\).