A block of mass 0.5 kg has an initial velocity of 10 `ms^(-1)` down an inclined plane of angle 309, the coefficient of friction between the block and the inclined surface is 0.2. The velocity of the block after it travels a distance of 10 m is (Take g = `10 ms^(-2)`)

Options

A. `17 ms^(-1) `
B. `13 ms^(-1)`
C. `24 ms^(-1)`
D. `8 ms^(-1)`

Answer

B

Explanation

Here ` m = 0.5 kg , u = 10 ms^(-1) , theta = 30 ^@ , mu = 0.2 ,s = 10 m`
acceleration down the plane
` a= g ( sin - mu cos theta ) = 10 ( sin 30^@ - 0.2 cos 30^@ ) = 3.268 `
from` , v^2 = u^2 + 2as = 10^2 + 2( 3.268) xx 10 = 165 .36`
` or v= sqrt( 165.36 ) = 13 ms ^(-1)`