Question

A wedge is moving rightwards on which a block of mass $10kg$ is place on it. Friction coefficient between the wedge and the block is $0.8$. [take $g=10m/s^2]$. Select correct alternative(s) among the following options.

• A. If wedge is moving with constant velocity then friction acting on block is $64N$
• B. If wedge is moving with constant velocity then acceleration of block is zero
• C. If wedge is moving with $a=2\left(\hat{i}\right)m/s^2$ then friction acting on block is $44N$
• D. If wedge is moving with $a=10\left(\hat{i}\right)m/s^2$ then friction is $20N$, downward on the block along the inclined

Answer 1

The correct option is C If wedge is moving with $a=2\left(\hat{i}\right)m/s^2$ then friction acting on block is $44N$

$C)$ if the wedge is moving with =$2\left(\hat{i}\right)m/s^2$ then friction acting on the block is 44N.

force due to gravity in the direction of incline will be

$Mg\times \sin 37=60$

if we go into the frame of wedge pseudo force acting on block will be $Ma$ where $a=2$ .Component of pseudo force along the incline will be

$Ma\times \cos 37=16N$

to hold the block at rest static friction will be

$60-16=44N$

maximum value of friction is

$\mu (Mg\cos \Theta +masin\Theta =76N$

here friction will not go upto its maximum value as it is not required and and the friction is static not kinetic.

So the friction force acting on the block to hold it will be $44N$