question_answer

A block of mass 20 kg is lying on a frictionless table. A block of 5 kg is kept on a block of 20 kg. If a variable force F given by \[F=kx\]is applied on the block of mass 20 kg and initially the mass of 20 kg is lying at \[x=1\text{ }m\]and \[\mu =0.2\] and \[k=5\text{ }N/m,\] find the distance after which 5 kg mass starts slipping from the starting point.

A) \[8\,m\]                  

B) \[3\,m\]                  

C) \[9\,m\]           

D) \[6\,m\]

Correct Answer: C

Solution :

[c] Let at \[x={{x}_{0}},\]5 kg starts slipping. \[F=k{{x}_{0}}=25a\] \[m=5kg,\,\,M=20kg\] \[a=\frac{5{{x}_{0}}}{25}=\frac{{{x}_{0}}}{5}\] \[\mu 5g=ma\] \[a=\mu g\] \[\frac{{{x}_{0}}}{5}=0.2\times 10\Rightarrow {{x}_{0}}=10m\] Distance from starting point is 9 m when mass of 5 kg starts slipping.