A) \[g(\cos \alpha -\mu \sin \alpha )\]
B) \[g(\sin \alpha -\mu cos\alpha )\]
C) \[\mu cos\alpha \]
D) \[g\sin \alpha \]
Correct Answer: B
Solution :
When a body descends, the frictional free acts upwards. The free body diagram given situation is as shown in the figure. When the body is moving down, frictional force acts upwards. Now the body descends under the action of the force mg. since\[-{{F}_{k}}\]. If downward acceleration is a, then, \[mg\sin \alpha -{{F}_{k}}=ma\] Where \[{{F}_{k}}=\mu R=\mu mg\cos \alpha \] \[\therefore \] \[mg\sin \alpha -\mu mg\cos \alpha =ma\] \[\Rightarrow \] \[a=g(\sin \alpha -\mu \cos \alpha )\]You need to login to perform this action.
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