Question

A block of mass 0.05 kg when placed on a rough 15${}^0$ inclined plane slides down without acceleration. The inclination is then increased to 30${}^0$. What would be the acceleration of the block?(g=9.8 ms${}^{-2}$)

Answer 1

Given,

$mass=0.05kg$

Now,

$f=ma$

$mgsin({15}^0+{30}^0)=ma$

$a=gsin({15}^0+{30}^0)$

$=g(sin{15}^{0}cos{30}^0+cos{15}^{0}sin{30}^2)$ Since $(sin{15}^0=0.643,cos{15}^0=0.965,cos{30}^0=\frac{\sqrt{3}}{2}=0.866)$

$=9.8(0.643\times 0.866+0.965\times 0.5)$

$9.8\left(1.039\right)$

$10.18m/s^2$

Thus the acceleration of the block is $10.18m/s^2$