Question

A board of mass $m=11kg$ is placed on the floor and a man of mass $m=70kg$ is standing on the board as shown (Case$-I$). The coefficient of friction between the board and the floor is $m=0.25$ If in Case$-II$ all surfaces are frictionless, the acceleration of man will be:

• A. None of these
• B. $\frac{2281}{291}\frac{m}{s^2}$
  
• C. $65\frac{m}{s^2}$
  
• D. $\frac{2800}{291}\frac{m}{s^2}$
  

Answer 1

The correct option is D $\frac{2800}{291}\frac{m}{s^2}$



Equation of motion of man
$Mg-T=Ma\dots \left(i\right)$
For board
From constrained relation
$A\frac{a}{2}$
$2T=mA=m\left(\frac{a}{2}\right)\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$a=\frac{4mg}{(4M+m)}$
$a=\frac{4\times 70\times 10}{(4\times 70+11)}$
$=\frac{2800}{291}\frac{m}{s^2}$

Final Answer: Option(a)