Question

A block of mass $1kg$ is stationary with respect to a conveyer belt that is accelerating with $1m/s^2$ upwards at an angle of ${30}^{\circ }$ as shown in figure. Which of the following is/are correct?

• A. Force of friction on block is $6N$ upwards along the inclined plane
• B. Force of friction on block is $1.5N$ upwards along the inclined plane
• C. Contact force between the block and belt is $10.5N$
• D. Contact force between the block and belt is $5\sqrt{3}N$

Answer 1

Answer: (A), (C)

The correct options are
A Force of friction on block is $6N$ upwards along the inclined plane
C Contact force between the block and belt is $10.5N$

Block F.B.D

$\theta ={30}^o$
$a=1m{s}^{-2}$

$g=10m{s}^{-2}$

$fr-mg\sin \theta =ma$

$fr=m(a+gsin\theta$

$fr=1\times (1+10\times \frac{1}{2}$)

$=1+5=6N$

$fr=6N$

Contact force =$\sqrt{(fr{)}^2+(\text{Normal}r\times n{)}^2}$

$=\sqrt{6^2+(mgcos\theta {)}^2}$

$\sqrt{6^2+(10\times \frac{\sqrt{3}}{2}{)}^2}$

$=\sqrt{36+75}$

$=\sqrt{111}$

$=10.5N$

Answer= A,C.