Dabour Hub EGEgyptian Curriculum 2026/2027
2-1 (g) Apparent weight with changing acceleration

2-1 (g) Apparent weight with changing acceleration

Point 7 of 7 — Motion with Variable Acceleration and ForceQuestion bank: 0 questions (0 with pictures)

07 2-1 (g) Apparent weight with changing acceleration

Idea

  • A person in a lift presses on the floor with force N=m(g+a) when the lift accelerates upward with acceleration a.
  • If a varies with time, so does N; just replace a by the function.
✍️ Worked examples

Example 19:

  • A lift accelerates upwards with a=0.2t m/s2.
  • A man of mass 60 kg stands in it.
  • Find the pressure on the floor at t=5 s in kg·wt, and the speed of the lift at that time (starting from rest).

Solution:

  • At t=5: a=1 m/s2.
  • N=60(9.8+1)=648 N =6489.8≈66.1 kg·wt.
  • v=∫050.2t dt=0.1(25)=2.5 m/s.

Common mistakes

  • Using s=ut+12at2 when the acceleration is not constant.
  • Forgetting the constant of integration, or using the wrong initial condition.
  • Mixing the forms a=dvdt and a=vdvds.
  • Confusing displacement with distance when the velocity changes sign.
  • Mixing units (grams with newtons, or cm with m).

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