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WAEC 2023 · Paper 1 · Q22

The velocity of a body of mass 4.564.56 kg increases from (10 m s−1,060∘)(10\ \text{m s}^{-1}, 060^\circ) to (50 m s−1,060∘)(50\ \text{m s}^{-1}, 060^\circ) in 16 seconds. Calculate the magnitude of the force acting on it.

Worked solution (try it first)
  1. Both velocities have the same bearing, 060∘060^\circ, so the change in velocity is 50−10=40 m s−150 - 10 = 40\ \text{m s}^{-1} along that line.
  2. The acceleration is 4016=2.5 m s−2\dfrac{40}{16} = 2.5\ \text{m s}^{-2}.
  3. By F=maF = ma, F=4.56×2.5=11.4F = 4.56 \times 2.5 = 11.4 N, option C.

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