Question

12.
In the arrangement shown in figure, coefficient of friction between the two blocks is u=o.5. The
force of friction acting between the two blocks is:

(1) 8N
(2) 10 N
(3) 6N
(4) 4N​

Answer 1

fbd is given in figure.let acceleration of both blocks towards left be a then

a=f-2/2=20-f/4

2f-4=20-f

f=8N

maximum friction between two blocks can be

fmax=umg(m=2kg)

       =(0.5)(2)(10)=10N

now,f<fmax

therefore friction force between two blocks is 8N

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Answer 2

Answer:

The frictional force acting between the two blocks is 8 N.

(1) is correct option.

Explanation:

Given that,

Coefficient of friction = 0.5

Force $F_{1}=2\ N$

Force $F_{2}=20\ n$

We need to calculate the net force

Using formula of force

$F=F_{2}-F_{1}$

Put the value into the formula

$F=20-2=18\ N$

We need to calculate the acceleration

Using formula of force

$F=ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{18}{6}$

$a=3\ m/s^2$

We need to calculate the frictional force acting between the two blocks

Using formula of frictional force

$f-F_{1}=ma$

$f=ma+F_{1}$

Put the value into the formula

$f=2\times3+2$

$f=8\ N$

Hence, The frictional force acting between the two blocks is 8 N.