Question

A block of $10kg$ mass is placed on a rough inclined surface as shown in figure. The acceleration of the block will be

Answer 1

Given,

Mass of the block $\left(m\right)=10kg$

Coefficient of static friction $\left(\mu s\right)=1.0$

Coefficient of kinetic friction $\left(\mu k\right)=0.8$

The angle between the inclined plane and horizontal $\left(\theta \right)={30}^{\circ }$

The Weight of the block $=mg$

Here, $g$ is the acceleration due to gravity.

Normal force $\left(N\right)=mg\cos \theta$

Net force acting on the inclined plane in the upward direction $\left(F\right)=(\mu s+\mu k)mg\cos \theta$

The Net force acting on the inclined plane in the downward direction $\left(f\right)=mg\sin \theta$

Let ${}^{\prime }{a}^{\prime }$ be the required acceleration of the block working downward then,

The equation of motion can be written as $f+ma=F$

$\Rightarrow mg\sin \theta +ma=(\mu s+\mu k)mg\cos \theta$

$\Rightarrow a=(\mu s+\mu k)g\cos \theta -g\sin \theta$

$\Rightarrow a=(1.0+0.8)gcos{30}^0-g\sin {30}^0$

$\Rightarrow a=1.56g-0.5g$

$\Rightarrow a=1.06g=1.06\times 9.8m/s^2=10.388m/s^2=10.4m/s^2$

Hence, the required acceleration will be $10.4m/s^2$