Question

As shown in the figure, A is a man of mass 60 kg standing on a block of mass 40 kg kept on ground. The coefficient of friction between the feet of the man and the block is 0.3 and that between B and ground is 0.2. If the person pulls the string with 125 N force, then

• A. B will slide on ground
• B. A and B will move with acceleration 0.5 ms${}^{-2}$
• C. the force of friction acting between A and B will be 40 N
• D. the force of friction acting between A and B will be 180 N

Answer 1

The correct option is D the force of friction acting between A and B will be 180 N

Lets start with calculating frictional forces.

Frictional force between man and block

$f_1=\mu m_{A}g$

$⟹f_1=$ 0.3 x 60 x 10

$⟹f_1=$ 180 N

Frictional force between block and ground

$f_2=\mu (m_A+m_B)g$

$⟹f_2=$ 0.2 x ( 60 + 40) x 10

$⟹f_2=$ 200 N

Now, total friction force on the block

$f_B=f_1+f_2$

$⟹f_B=$ 380 N

Tension in the string $T=125N$

Since, frictions forces on the block is more than the force applied by the man. Therefore, there will not be any motion.

Friction acting between A and B $=f_1=$$180N$