Question

A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to the upper block. Take g = 10 $m/s^2$.

• A. a = 4 m/s,a = 4 m/s
• B. a = 1 m/s,a = 4 m/s
• C. a =2 m/s,a =2 m/s
• D. None of these

Answer 1

The correct option is D

None of these

Step 1: to check if sliding is happening. Assume sliding is not happening, and they are moving with an acceleration a,
Block A,$\sum f_x=m_{A}a$
$\Rightarrow 12-f=m_{A}a......\left(1\right)$
Block B,$\sum f_x=m_{A}a$
$\Rightarrow f=m_{A}a.....\left(2\right)$
Adding (1)and (2)
$12=(m_A+m_A)a$
$\Rightarrow 12=(2+4)a$
$12=6a$
$\Rightarrow a=2m/s^2$
From (a)
f=4$\times$ 2 =8N
but $f_{max}$=$\mu \times N$
$\therefore$ For them to move together without slipping, friction force required is 8N but maximum value of friction is 4N. $\therefore$ our assumption, that, they do not slip is wrong, and f=$f_{mass}$=4N
$\therefore$ For block A,
$12-f=\frac{m_A}{a_A}$
$\Rightarrow 12-4=2a_A$
$\Rightarrow 8=2a_A$
$\Rightarrow a_A=4m/s^2$
Similarly for block,B,
$f=m_B{a}_B$
$\Rightarrow$ 4=4$a_B$ $\Rightarrow$ $a_B=1m/s^2$

From (a)

f = 4 $\times$ 2 = 8 N

but f${}_{max}$ = $\mu$$\times$ N

= 0.2 $\times$ 20 = 4 N

$\therefore$For them to move together without slipping, friction force
required is 8N but maximum value of friction is 4N. $\therefore$our
assumption, that, they do not slip is wrong,and f = f${}_{max}$ = 4N

$\therefore$For block A,

12 - f = m${}_A$/a${}_A$

$\Rightarrow$ 12 - 4 = 2 a${}_A$

$\Rightarrow$ 8 = 2 a${}_A$

$\Rightarrow$a${}_A$ = 4 m/s${}^2$

Similarly for block, B,

f = m${}_B$a${}_B$

$\Rightarrow$ 4 = 4 a${}_B$ $\Rightarrow$a${}_B$ = 1
m/s${}^2$