Question

The system starts from rest and block $A$ attains a velocity of $5\text{m/s}$ after it has moved $5\text{m}$ towards the right. Assuming the arrangement to be frictionless everywhere and pulley & strings to be light, find the value of the constant force $F$ applied on block $A$. (Take $g=10m/s^2$)

• A. $50\text{N}$
• B. $75\text{N}$
• C. $100\text{N}$
• D. $96\text{N}$

Answer 1

The correct option is B $75\text{N}$

The given system can be reframed as shown


From the constrained condition, using method of intercept we get
${\stackrel{..}{L}}_1+{\stackrel{..}{L}}_2+{\stackrel{..}{L}}_3=0$
$\Rightarrow a_A+a_A+0-a_B=0$
$\Rightarrow a_B=2a_A$...........(1)
Since A starts from rest, using 3rd equation of motion, we get
$a_A=\frac{v^2}{2s}=\frac{5^2}{2\times 5}=2.5m/s^2$
$\Rightarrow a_B=5m/s^2$

From the FBDs of the masses $6kg$ and $2kg$, we have
$T-2g=2a_B\Rightarrow T=2g+2a_B=20+2\times 5=30N$
$F-2T=6a_A\Rightarrow F=2T+6a_A=2\times 30+6\times 2.5=75N$
Thus, the value of $F$ is $75N$