A block is kept on a frictionless inclin...

A block is kept on a frictionless inclined surface with angle of inclination `alpha`. The incline is given an acceleration 'a' to keep the block stationary. Then a is equal to

`g tanalpha`

`(g)//(tanalpha)`

`gcosec alpha`

Text Solution

Verified by Experts

The correct Answer is:

`uarr:Ncosalpha=mgrArrN=mg//cosalpha`
`rarr:Nsinalpha=marArr(mgsinalpha)/(cosalpha)=ma`
`a=g tanalpha`

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