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Excellent! Force is change in momentum divided by the time taken. Momentum \(\vec{p} = m \vec{v}\). So, we subtract the initial momentum from the final momentum, and divide this by the time taken.
The unit of Force is Newton (N). \(1 \, \mathrm{N} = 1 \, \mathrm{kg} \, \mathrm{m/s^2}\).
If the \(500 \, \mathrm{kg}\) lorry's velocity decreases from \(8 \, \mathrm{m/s}\) to \(4 \, \mathrm{m/s}\) in \(5 \, \mathrm{s}\), what would be average Force acting on it?
  • \(\frac{\left(500 \times 5 \right) - \left(500 \times 8 \right)}{4}\)
  • \(\frac{\left(500 \times 8 \right) - \left(500 \times 5 \right)}{4}\)
  • \(\frac{\left(500 \times 4 \right) - \left(500 \times 8 \right)}{5}\)
  • \(\frac{\left(500 \times 8 \right) - \left(500 \times 4 \right)}{5}\)