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A particle of mass mm is driven by a machine that delivers a constant power of kk watts. If the particle starts from rest, the force on the particle at time tt is:

A

mk2t−1/2\sqrt{\frac{mk}{2}}t^{-1/2}

B

mkt−1/2\sqrt{mk}t^{-1/2}

C

2mkt−1/2\sqrt{2mk}t^{-1/2}

D

12mkt−1/2\frac{1}{2}\sqrt{mk}t^{-1/2}

Step-by-Step Solution

  1. Relation between Power, Force, and Velocity: Power (PP) is defined as the product of force (FF) and velocity (vv). Given constant power P=kP = k: P=F⋅v=kP = F \cdot v = k
  2. Express Force in terms of Velocity: Using Newton's Second Law, F=ma=mdvdtF = ma = m \frac{dv}{dt}. Substituting this into the power equation: mdvdt⋅v=km \frac{dv}{dt} \cdot v = k v dv=kmdtv \, dv = \frac{k}{m} dt
  3. Integrate to find Velocity: Integrate both sides from initial state (v=0,t=0v=0, t=0) to state at time tt: ∫0vv dv=km∫0tdt\int_{0}^{v} v \, dv = \frac{k}{m} \int_{0}^{t} dt v22=kmt  ⟹  v=2kmt1/2\frac{v^2}{2} = \frac{k}{m} t \implies v = \sqrt{\frac{2k}{m}} t^{1/2}
  4. Calculate Acceleration: Differentiate velocity with respect to time to find acceleration (aa): a=dvdt=ddt(2kmt1/2)a = \frac{dv}{dt} = \frac{d}{dt} \left( \sqrt{\frac{2k}{m}} t^{1/2} \right) a=2km⋅12t−1/2a = \sqrt{\frac{2k}{m}} \cdot \frac{1}{2} t^{-1/2}
  5. Calculate Force: Finally, calculate force F=maF = ma: F=m(2km⋅12t−1/2)=m22kmt−1/2F = m \left( \sqrt{\frac{2k}{m}} \cdot \frac{1}{2} t^{-1/2} \right) = \frac{m}{2} \sqrt{\frac{2k}{m}} t^{-1/2} F=m24⋅2kmt−1/2=mk2t−1/2F = \sqrt{\frac{m^2}{4} \cdot \frac{2k}{m}} t^{-1/2} = \sqrt{\frac{mk}{2}} t^{-1/2}
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