A) 11.9 N
B) 1.19 N
C) 0.19 N
D) 1109 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, \[=R=mg\text{ }cos30{}^\circ \] ... (ii) From Eqs. (i) and (ii), we get \[F=\mu \text{ }mg\text{ }cos\text{ }30{}^\circ \] Given, \[\mu =0.7,\text{ }m=2\text{ }kg,\text{ }g=9.8\text{ }m/{{s}^{2}},\] \[\cos {{30}^{o}}=\frac{\sqrt{3}}{2}\] \[\therefore \] \[F=0.7\times 2\times 9.8\times \frac{\sqrt{3}}{2}=11.9N\]You need to login to perform this action.
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