JEE Advanced 2021 Paper 1 · Physics · Kinematics and Laws of Motion · medium

A block A of mass 2 kg rests on a block B of mass 4 kg, and B lies on a smooth horizontal floor. The coefficient of friction between A and B is 0.30…

One or more options correct

A block

A

of mass

2\ \mathrm{kg}

rests on a block

B

of mass

4\ \mathrm{kg}

, and

B

lies on a smooth horizontal floor. The coefficient of friction between

A

and

B

0.30

(static equal to kinetic). A horizontal force

F

is applied to

A

. Take

g=10\ \mathrm{m\,s^{-2}}

. Choose the correct statement(s).

A. For $F=6\ \mathrm{N}$ there is no slipping and the friction force on $B$ is $4\ \mathrm{N}$ .
B. For $F=12\ \mathrm{N}$ the accelerations of $A$ and $B$ are $3.0\ \mathrm{m\,s^{-2}}$ and $1.5\ \mathrm{m\,s^{-2}}$ respectively.
C. Slipping between $A$ and $B$ begins as soon as $F$ exceeds $6\ \mathrm{N}$ .
D. For $F=12\ \mathrm{N}$ , $A$ slides $3.0\ \mathrm{m}$ relative to $B$ in the first $2\ \mathrm{s}$ after $F$ is applied (starting from rest).

Show answer and solution

Answer: A, B, D

The maximum friction is

f_{max}=\mu m_Ag=0.3\times2\times10=6

N. (A) Without slipping, both move with

a=F/6

;

B

is accelerated only by friction, so

f=4a=4F/6

. For

F=6

N this gives

a=1\ \mathrm{m\,s^{-2}}

and

f=4

N, which is below

6

N: correct. (C) Slipping starts when

4F/6=6

, i.e.

F=9

N, not

6

N: incorrect. (B) At

F=12

N slipping occurs;

A

a_A=(12-6)/2=3.0\ \mathrm{m\,s^{-2}}

B

a_B=6/4=1.5\ \mathrm{m\,s^{-2}}

: correct. (D) The relative acceleration is

1.5\ \mathrm{m\,s^{-2}}

, so the relative displacement in

2

s is

\tfrac12\times1.5\times4=3.0

m: correct. Hence (A), (B) and (D).

Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake