A) 2.5 N
B) b) 5N
C) 7.84 N
D) 10N
Correct Answer: A
Solution :
Limiting frictional force \[{{F}_{s}}={{\mu }_{s}}R\] \[={{\mu }_{s}}mg=0.4\times 2\times 9.8=7.84N\] Applied force \[2.5\text{ }N<7.84\text{ }N\] So, block cannot move and force of friction will be equal to applied force. Force of friction\[=2.5\text{ }N\].You need to login to perform this action.
You will be redirected in
3 sec