A) \[Ty\]
B) \[\frac{\sqrt{T}}{y}\]
C) \[\frac{T}{y}\]
D) \[\frac{T}{\sqrt{y}}\]
Correct Answer: C
Solution :
A block of mass m is moving initially with velocity u on a rough surface and due to friction, it comes to rest after covering a distance s Retarding force \[ma=\mu \,\,mg\] \[\therefore \] \[a=\mu g\] From \[{{v}^{2}}={{u}^{2}}-2as\] \[\Rightarrow \] \[0={{u}^{2}}-2\mu gs\] \[\frac{2T}{m}=2\mu gy\] ?..(i) \[\left[ As\,\,T=\frac{1}{2}m{{u}^{2}} \right]\] Now, frictional force / =p, mg ...(ii) ie, from Eqs. (i) and (ii), we get \[f=\frac{T}{y}\]
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