Quantities and units in mechanics

M6 Quantities and units in mechanics — coverage pack

1 specification leaves · notes, questions, answers and worked methods

M6.1 · Understand and use fundamental quantities and units in the S.I. system: length, time, mass; understand and use derived quantities and units: velocity, acceleration, force, weight, moment.

  • The fundamental S.I. quantities are length in metres, time in seconds and mass in kilograms; velocity uses m s−1\text{m s}^{-1}, acceleration m s−2\text{m s}^{-2}, force and weight newtons, and moment N m\text{N m}.
  • Convert all measurements to consistent S.I. units before substituting, using for example 1 km h−1=518 m s−11\,\text{km h}^{-1}=\dfrac{5}{18}\,\text{m s}^{-1} and 1 tonne=1000 kg1\,\text{tonne}=1000\,\text{kg}.
  • Since F=maF=ma, 1 N=1 kg m s−21\,\text{N}=1\,\text{kg m s}^{-2}; multiplying a force by a perpendicular distance gives a moment in N m\text{N m}.
  • A common error is to confuse mass with weight: mass is measured in kilograms, whereas weight is a force measured in newtons; a moment is not measured in newtons alone.

Tier 1 · Easy

  1. 1. Convert 90 km h−190\,\text{km h}^{-1} to metres per second.[2 marks]

    Answer

    • 25 m s−125\,\text{m s}^{-1}

    Method: 90 km h−1=90×10003600 m s−1=90×518 m s−1=25 m s−190\,\text{km h}^{-1}=90\times\dfrac{1000}{3600}\,\text{m s}^{-1}=90\times\dfrac{5}{18}\,\text{m s}^{-1}=25\,\text{m s}^{-1}.

Tier 2 · Standard

  1. 1. A resultant force gives a 750 kg750\,\text{kg} car an acceleration of 1.6 m s−21.6\,\text{m s}^{-2}. Find the force in newtons and write the newton in fundamental S.I. units.[3 marks]

    Answer

    • 1200 N1200\,\text{N}
    • 1 N=1 kg m s−21\,\text{N}=1\,\text{kg m s}^{-2}

    Method: Using F=maF=ma, F=750×1.6=1200 NF=750\times1.6=1200\,\text{N}. Since mass has unit kg\text{kg} and acceleration has unit m s−2\text{m s}^{-2}, the fundamental-unit form is kg m s−2\text{kg m s}^{-2}.

Tier 3 · Hard

  1. 1. A vehicle of mass 1.81.8 tonnes increases its speed uniformly from 54 km h−154\,\text{km h}^{-1} to 90 km h−190\,\text{km h}^{-1} in 8.0 s8.0\,\text{s}. Find the resultant force. This force has a perpendicular distance of 0.65 m0.65\,\text{m} from a pivot; find its moment about the pivot.[6 marks]

    Answer

    • 2250 N2250\,\text{N}
    • 1462.5 N m1462.5\,\text{N m}

    Method: The mass is 1800 kg1800\,\text{kg}. The speeds are 54×5/18=15 m s−154\times5/18=15\,\text{m s}^{-1} and 90×5/18=25 m s−190\times5/18=25\,\text{m s}^{-1}, so a=(25−15)/8=1.25 m s−2a=(25-15)/8=1.25\,\text{m s}^{-2}. Hence F=ma=1800(1.25)=2250 NF=ma=1800(1.25)=2250\,\text{N}. The moment is Fd=2250(0.65)=1462.5 N mF d=2250(0.65)=1462.5\,\text{N m}.