As shown in the figure M is a man of mass 60 kg

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As shown in the figure M is a man of mass 60 kg

 

 

Question :- As shown in the figure, M 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 the ground is 0.1 . If the person pulls the string with 100 N force, then :

(A) B will slide on ground

(B) A and B will move together with acceleration 1 m/s2

(C) the friction force acting between A & B may be 40 N

(D) the friction force acting between A & B may be 180 N

As shown in the figure M is a man of mass 60 kg - Curio Physics

 

Solution :-

Limiting friction between feet of man and block B :-

\displaystyle {{f}_{lim}}=\mu N=\mu {{m}_{A}}g=0.3\times 60\times 10=180N

As the man pulls the rope with 100 N force towards left, the same 100 N force acts on him towards right. But to move the person w.r.t. block, at least 180 N force is required. Hence the person will not move with resect to block B.

Now let us assume that the block B moves with acceleration ‘a’ on the ground and write equations of motion of man A and block B and their FBD :-

As shown in the figure M is a man of mass 60 kg - Curio Physics

For man :-

\displaystyle 100-{{f}_{s}}=60\times a     …..(1)

For block B :-

\displaystyle 100+{{f}_{s}}-{{f}_{k}}=40\times a     …..(2)

As

\displaystyle {{f}_{k}}=0.1\times (60+40)\times 10=100N

Hence from equation (2),

\displaystyle 100+{{f}_{s}}-100=40\times a

\displaystyle \Rightarrow {{f}_{s}}=40\times a     …..(3)

Using equations (1) and (3), we get

\displaystyle 100-40\times a=60\times a

\displaystyle \Rightarrow 100a=100

\displaystyle \Rightarrow a=1m{{s}^{-2}}

From equation (3),

\displaystyle {{f}_{s}}=40\times a=40\times 1=40N

Hence options (A), (B) and (C) are correct.

Questions, Class 11th NCERT
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