Question

A block rests on a rough inclined plane making an angle of $3^o$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10N$, the mass of the block in (kg) is: (Take $g=10m{s}^{-2}$)

• A. $2.0$
• B. $4.0$
• C. $1.6$
• D. $2.5$

Answer 1

The correct option is A $2.0$

$\text{Step 1: Draw free body diagram and Resolving forces}$ $\text{[Ref. Fig.]}$

Let $f$ be the frictional force

As shown in the figure:

Resolving force $mg$ along the inclined plane, we get $mgsin\theta$

Resolving force $mg$ perpendicular to the incline, we get $mgcos\theta$

$\text{Step 2: Net force on block}$

Since the block is at rest, therefore net force acting on it along inclined direction is zero.

$\sum F_{net}=0$

$\Rightarrow f-mg\sin \theta =0$

$\Rightarrow$ $10=m\times 10\times \sin {30}^0$

$\Rightarrow$ $m=2kg$

Hence, Option $A$ is correct.

$\text{Note:}$ $\text{There is no use of friction coefficient in this question.}$

Since the block is not slipping and the value of static friction is given. So, this value must be less than or equal to its maximum value $\left(\mu N\right)$. Hence we need not compare it.