When an object slides at a constant speed down an inclined plane, the coefficient of friction may approximately be:
Correct answer: C. Tan∅
- A. sin∅
- B. Cos∅
- C. Tan∅
- D. cot∅
Explanation
We know,f =uN .........(1)where u=coefficient of kinetic frictionN = normal forceN =mg cos ∅ (as in the diagram, N balances mgcos∅)and f= mg sin∅ (as the body moves with constant speed, so f balances mgsin∅)Put these in eqn (1)mgsin∅=u×mgcos∅u =sin∅/cos∅u = tan∅
This question appeared on the ETEA MDCAT 2005 paper, which you can sit online with every answer explained.
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