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Derive the fundamental SI units of power using the definitions of power, work, force, and acceleration.

Physics

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Answer

kg m2/s3\text{kg m}^2/\text{s}^3

Step-by-step solution

5 steps
  1. Define power in terms of work and time.

    Power (PP) is the rate at which work (WW) is done or energy is transferred over time (tt).

    P=WtP = \frac{W}{t}
  2. Define work in terms of force and displacement.

    Work (WW) is the product of force (FF) and displacement (dd).

    W=F⋅dW = F \cdot d

    Substitute this into the power equation:

    P=F⋅dtP = \frac{F \cdot d}{t}
  3. Define force in terms of mass and acceleration.

    Force (FF) is the product of mass (mm) and acceleration (aa).

    F=m⋅aF = m \cdot a

    Substitute this into the power equation:

    P=m⋅a⋅dtP = \frac{m \cdot a \cdot d}{t}
  4. Express the fundamental SI units for each quantity.

    • Mass (mm): kg\text{kg}
    • Acceleration (aa): m/s2\text{m/s}^2 (since acceleration is change in velocity over time, and velocity is change in displacement over time, a=m/ss=m/s2a = \frac{\text{m/s}}{\text{s}} = \text{m/s}^2)
    • Displacement (dd): m\text{m}
    • Time (tt): s\text{s}
  5. Substitute the fundamental units into the expression for power.

    Punits=(kg)⋅(m/s2)⋅(m)(s)P_{\text{units}} = \frac{(\text{kg}) \cdot (\text{m/s}^2) \cdot (\text{m})}{(\text{s})} Punits=kg⋅m2s2⋅sP_{\text{units}} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{s}} Punits=kg⋅m2s3P_{\text{units}} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} Punits=kg m2s−3P_{\text{units}} = \text{kg m}^2 \text{s}^{-3}

    Thus, the fundamental units of power are kg m2/s3\text{kg m}^2/\text{s}^3.

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