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A block of mass 15 kg is lying on an inc...

A block of mass 15 kg is lying on an inclined plane of angle `20^(@)`. In order to make it move upward along the slope with an acceleration of 25 cm / `s^(2)` a horizontal force of 200 N is required to be applied on it. Calculate (i) frictional force on the block and (ii) co-efficient of kinetic friction.

Text Solution

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The situatin described here is shown in figure we take X-axis in the direction parallel to the surface of the slope and so Y-axis would be in Remember that the block performs acceleration force on it is not zero .
F `cos 20^(@) -f- mg sin 20^(@) = (15) (0.25)`
Since the block is in equilbrium in the y-axis
`Sigma F_(y)=0`
N- mg `cos 20^(@) - F sin 20^(@) =0`
`N= 138. 1359 - 68.4= 0`
`N= 297 N`
`mu_(k) = (f )/( N) = (134 )/( 207) = 0.65`
Note here that due tothe force F the normal force increase.
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