Question

Three blocks of masses $m_1, m_2$ and $m_3$ are connected by massless strings as shown on a friction less table. They are pulled with a force of 40 N if $m_1$ = 10 kg, $m_2$ = 60 kg and $m_3$ = 4 kg then tension $T_2$ will be

• A. 10 N
• B. 20 N
• C. 32 N
• D. 40 N

Answer 1

Answer: (C)

32 N

Answer 2

Since, the table is friction less ie, it is smooth therefore, force on the blocks is given by $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, F = (m_1 + m_2 + m_3)a$ $\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, a= \frac{F}{m_1 + m_2 + m_3}$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{40}{10 + 6 + 4} = \frac{40}{20}$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 2ms^{-2}$ Now the tension between 10 kg and 6 kg masses is given by $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, T_2 = (m_1 + m_2)a$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = (10 + 6 ) 2 = 16 \times 2$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 32 N$