Question

Three masses $M=100kg,m_1=10kg$ and $m_2=20kg$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $F$ is applied on the system so that the mass $m_2$ moves upward with an acceleration of $2m{s}^{-2}$. The value of $F$ is :
$($ Take $g=10m{s}^{-2})$

• A. 3360 N
• B. 3380 N
• C. 3120 N
• D. 3240 N

Answer 1

Answer: (A)

Let acceleration of $100kg$ block $=a_1$
FBD of $100kg$ block w.r.t ground

$F-T-N_1=100a_1$
FBD of $20$ block wrt $100kg$

$T-20g=20(2)$
$T=\mathrm{240.....}$(ii)
$N_1=20a_1\dots \text{.... (iii) }$
FBD of $10kg$ block wrt $100kg$

$a_1=26m/s^2$
$F-240-20(26)=100\times 26$
$\Rightarrow F=3360N$