CBSE-XI-Physics
05: Newton's Laws of Motion
- #11A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially, the block B is near the right end of block A (Figure 5-E3). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time that elapses before block B separates from A.
Figure
Ans : Mass of block A, m = 5 kg
F = ma = 10 N
`` \Rightarrow a=\frac{10}{5}=2\,\mathrm{\,m \,}/{\,\mathrm{\,s \,}}^{2}``
As there is no friction between A and B, when block A moves, block B remains at rest in its position.
Initial velocity of A, u = 0
Distance covered by A to separate out,
s = 0.2 m
Using `` s=ut+\frac{1}{2}a{t}^{2}``, we get:
`` 0.2=0+\frac{1}{2}\times 2{t}^{2}``
⇒ t2 = 0.2
⇒ t = 0.44 s ≈ 0.45 s
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