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 \[ma=\mu \text{ }mg\] \[\therefore \] \[a=\mu g\] From \[{{v}^{2}}={{u}^{2}}-2as\] \[\Rightarrow \] \[0={{u}^{2}}-2\mu g\] \[\frac{2T}{m}=2\mu gy\] ...(i) \[\left[ As\,T=\frac{1}{2}m{{u}^{2}} \right]\] Now, frictional force \[f=\mu \,mg\] ... (ii) ie, from Eqs. (i) and (ii), we get \[f=\frac{T}{y}\]
You need to login to perform this action.
You will be redirected in
3 sec
