A) \[11.9\text{ }N\]
B) \[1.19N\]
C) \[0.19\text{ }N\]
D) \[11.0\text{ }N\]
Correct Answer: A
Solution :
The various forces acting on the block are as shown Frictional force \[=\mu R\] ...(i) where \[\mu \] is coefficient of friction and R the normal reaction of the surface on the block. Also, \[R=mg\,\,\cos {{30}^{o}}\] ... (ii) From Eqs. (i) and (ii), we get \[F=\mu \,mg\,\cos \,{{30}^{o}}\] Given, \[\mu =0.7,\,\,m=2kg,\,g=9.8m/{{s}^{2}},\] \[\cos \,{{30}^{o}}=\frac{\sqrt{3}}{2}\] \[\therefore \] \[F=0.7\times 2\times 9.8\times \frac{\sqrt{3}}{2}=11.9N\]You need to login to perform this action.
You will be redirected in
3 sec