A) 2.0
B) 4.0
C) 1.6
D) 2.5
Correct Answer: A
Solution :
Let the mass of block be\[m\]. Frictional force in rest position \[F=mg\sin 30{}^\circ \] [This is static frictional force and may be less than the limiting frictional force] \[10=m\times 10\times \frac{1}{2}\] \[\therefore \] \[m=\frac{2\times 10}{10}=2\,kg\]You need to login to perform this action.
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