A) 2.0 kg
B) 4.0 kg
C) 1.6kg
D) 2.5kg
Correct Answer: A
Solution :
As the block is at rest, so the frictional force must be equal to the component of the weight along the inclined plane. \[\therefore \]\[mg\,\sin \theta =10\Rightarrow m\times 10\times \frac{1}{2}=10\Rightarrow m=2\,kg\] Hence, the correction option is [a].You need to login to perform this action.
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