A block of mass 2 kg rests on an inclined plane which makes an angle of `30^(@)` with the horizontal. The coefficient of friction between the block and the surfcae is `sqrt(3//2)`. (i) What force should be applied on the block so that it moves down without any acceleration ? (ii) What force should be applied on the block so that it moves up without any acceleration ? (iii) Calculate the ratio of the powers in the above two cases if the block moves with uniform speed in both the cases.

Explanation

Here `f_(L) = muR = mu "mg cos" theta `
`= 3sqrt2 xx 2 xx g xx (sqrt3//2) = 3g//sqrt2` and component of weight along the plane
`"mg sin" theta = 2 xx g xx (1//2) = g`
(a) As `"mg sin" theta lt f_(L)` , the block will remain at rest on the plane and so to move it down the plane force F must be applied parallel to the plane downwards as shown in Fig. 7.43 (a) .
So , equation of motion down the plane will be

`(F_(D) + "mg sin" theta) - f_(L) = "ma"`
or ` " " F_(D) = 2g(1)/(2) = g(3)/(sqrt2) " " `[as v = constant so a , 0]
or `" " F_(D) = ((3-sqrt2)/(sqrt2))g = 11 N`
(b) If `F_(U)` is the force required to move the block up the plane , the equation of motion will be
`F_(U) - (f_(L) + "mg sin" theta ) = "ma" `
or `" " F_(U) = 2g(1)/(2) + g(3)/(sqrt2)` [ as v = constant so a = 0]
or `" " F_(U) = ((3 + sqrt2)/(sqrt2)) g = 30.6 N`
(c) As power = (force) `xx` (velocity)
So , `" " (P_(D))/(P_(U)) = (F_(D)v_(D))/(F_(U)v_(U)) = (F_(D))/(F_(U)) " " ["as" v_(D) = v_(U)]`
i.e., `" " (P_(D))/(P_(U)) = [(3 - sqrt2)/(3 + sqrt2)] = 0.36`