Two blocks of masses m-1-1 kg and m-2-2 kg are connected by massless strings and placed on a horizontal frictionless surface as shown in figure- A force F-12 N is applied to mass m-1- as shown then tension -T- between m-1- and m-2-4 N8 N10 N2 N

Question

Toppr Admin · Accepted Answer

Net force on the system-m1-m2-3kg- is F-12N so the common acceleration-xA0- will be a-F3kg-4m-s2

Two blocks of masses $m_{1} = 1 k g$ and $m_{2} = 2 k g$ are connected by massless strings and placed on a horizontal frictionless surface as shown in figure. A force $F = 12 N$ is applied to mass $m_{1}$ as shown then tension $(T)$ between $m_{1}$ and $m_{2}$

$4 N$
$8 N$
$10 N$
$2 N$

Net force on the system $(m_{1} + m_{2} = 3 k g)$ is $F = 12 N$ so the common acceleration will be $a = \frac{F}{3 k g} = 4 m / s^{2}$

Two blocks of masses m1=1 kg and m2=2 kg are connected by massless strings and placed on a horizontal frictionless surface as shown in figure. A force F=12 N is applied to mass m1 as shown then tension (T) between m1 and m24 N8 N10 N2 N

Net force on the system(m1+m2=3kg) is F=12N so the common acceleration will be a=F3kg=4m/s2

Two blocks of masses $m_{1} = 1 k g$ and $m_{2} = 2 k g$ are connected by massless strings and placed on a horizontal frictionless surface as shown in figure. A force $F = 12 N$ is applied to mass $m_{1}$ as shown then tension $(T)$ between $m_{1}$ and $m_{2}$

$4 N$
$8 N$
$10 N$
$2 N$

Net force on the system $(m_{1} + m_{2} = 3 k g)$ is $F = 12 N$ so the common acceleration will be $a = \frac{F}{3 k g} = 4 m / s^{2}$