Question

In the shown figure, a man on a horizontal platform $A$, kept on a smooth horizontal surface, holds a rope attached to a box $B$. Man pulls the rope with a constant force of $50\text{N}$. The co-efficient of friction between the man and the platform is $0.5$. Then (Mass of man is $80kg$, mass of platform is $120kg$ and mass of box is $100kg$)

• A. velocity of box relative to platform after $4s$ is $3m/s$.
• B. velocity of man relative to platform after $4s$ is $2m/s$.
• C. friction force between man and platform is $40N$.
• D. friction force between man and platform is $30N$.

Answer 1

Answer: (A), (D)

friction force between man and platform is $30N$.

The FBDs are as shown


From the FBD of man, we have
$N=80g=800\text{N}$
Maximum friction force between platform $A$ and man is
$f_{max}=\mu N=0.5\times 800=400\text{N}$
$\because 400N>50N$, there will be no slipping between platform and man.

So, taking man and platform as a system, the acceleration of the platform is given as
$a_A=\frac{50}{m_M+m_A}=\frac{50}{80+120}=0.25m/s^2$

From the FBD of box B, we have
acceleration of the box as
$a_B=\frac{50}{m_B}=\frac{50}{100}=0.5m/s^2$

So, relative acceleration is
$a_{BA}=a_B-a_A=\left(0.5\right)-(-0.25)=0.75m/s^2$
Thus, relative velocity after $t=4s$ will be given by
$v_{BA}=u_{BA}+a_{BA}t$
$\Rightarrow v_{BA}=0+0.75\times 4=3m/s$
Also, friction between man and platform $A$ is
$f_s=m_A\times a_A=120\times 0.25=30\text{N}$