CBSE-XI-Physics

05: Newton's Laws of Motion

with Solutions - page 5
Qstn# iv-13 Prvs-QstnNext-Qstn
  • #13
    The elevator shown in figure (5-E5) is descending with an acceleration of 2 m/s2. The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B?
    Figure
    Ans : When the elevator is descending, a pseudo-force acts on it in the upward direction, as shown in the figure.

    From the free-body diagram of block A,
    `` mg-N=ma``
    `` N=m\left(g-a\right)``
    `` \Rightarrow N=0.5\left(10-2\right)=4\,\mathrm{\,N \,}``
    `` ``
    So, the force exerted by the block A on the block B is 4 N.
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