Question

a block metal of mass 50g when placed over an inclined plane at an angle of 15° slides down without acceleration.If the inxlination is increased by 15°,what would be the acceleration of the block?

Answer 1

a block of metal of mass 50g when placed over an inclined plane at an angle of 15° slides down without acceleration. it means, block is moving in equilibrium condition .
e.g., downward force along plane = upward force along plane
mgsin15° = fr
we know, fr = $\mu$N
and N = mgcos15°
so, mgsin15° = $\mu$mgcos15°
$\mu=tan15^{\circ}$

now, when inclination increased by 15°
then, mgsin(15°+15°) - fr = ma
mgsin30° - $\mu$mgcos(15°+15°)=ma
gsin30° - tan15°.gcos30° = a
a = g{sin30° - tan15°.cos30°}
a = 10{0.5 - (2-1.73)0.86}
a = 10{0.5 - 0.27 × 0.86}
a = 2.678 m/s²

Answer 2

it's answer will be 2.69ms