Question

Two blocks of masses 5 kg. and 10 kg are connected through string and lies inside the cavity which is created in the wedge as shown in the figure, mass of the wedge is 25 kg and there is no friction between wedge and ground. Coefficient of friction between the block 5 kg and cavity (1) is 1/2. and between the block 10 kg and cavity (2) is 1/4. For what value of 'm' both block 5 kg and 10 kg is in rest w.r.t. wedge. $(g=10m/s^2)$:

• A. 40 kg
• B. 5 kg
• C. 10 kg
• D. Not possible
• E. Answer Required

Answer 1

The correct option is D Not possible

Let acceleration of wedge be a.

$N_1=m_{1}a$

$N_2=m_{2}a$

if block $m_1$ and $m_2$ is at rest w.r.t wedge, then;

$T-m_{1}g-F_2=0$ ...............(1)

and,

$M_{2}g-F_2-T=0$ ...............(2)

where $F_1$ and $F_2$ are the frictional forces;

$F_1={\mu }_1{N}_1=\frac{1}{2}{m}_{1}a$ $=\frac{5}{2}a$

$F_2={\mu }_2{N}_2=\frac{1}{4}{m}_{2}a$ $=\frac{5}{2}a$

From eq. (1) and (2);

$F_1-F_2=0$

$\Rightarrow 10\times 10-5\times 10-\frac{5}{2}a$ $-\frac{5}{2}a=0$

$\Rightarrow 50-5a=0$

$\Rightarrow 5a=50$

$\Rightarrow a=10m/s^2$

$mg-T=ma$ ...............(3)

$T=(m_1+m_2+m_3)a$

$T=(5+10+25)\times 10=40\times 10=400N$

Putting the value of T in eq. (3)

$mg-400=ma$

$m(g-a)=400$

$\Rightarrow m\times 0=400$

This is not possible.