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CBSE-XI-Physics

05: Newton's Laws of Motion

with Solutions - page 5
Qstn# iv-15-a Prvs-QstnNext-Qstn
  • #15-a
    the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s2.
    Figure
    Ans : Maximum weight will be recorded when the elevator accelerates upwards.
    Let N be the normal reaction on the person by the weighing machine.
    So, from the free-body diagram of the person,
    N=mg+ma ...(1)
    This is maximum weight, N = 72 × 9.9 N
    When decelerating upwards, minimum weight will be recorded.
    N’=mg+m-a ...(2)
    This is minimum weight, N‘ = 60 × 9.9 N
    From equations (1) and (2), we have:
    2 mg = 1306.8
    ⇒m=1306.82×9.9=66 kg
    So, the true mass of the man is 66 kg.
    And true weight = 66
    ×9.9 = 653.4 N (b) Using equation (1) to find the acceleration, we get:
    mg + ma = 72 × 9.9
    ⇒a=72×9.9-66×9.966=9.9×666=9.911⇒a=0.9 m/s2
  • #15-b
    the magnitude of the acceleration. Take g = 9.9 m/s2.
    Figure
    Ans : Using equation (1) to find the acceleration, we get:
    mg + ma = 72 × 9.9
    ⇒a=72×9.9-66×9.966=9.9×666=9.911⇒a=0.9 m/s2