Question

Two masses of 5 kg and 3 kg are suspended with the help of massless inextensible string as shown in figure . When whole system is going upwards with acceleration \(2 m/s^2\) . The value of T1 is ( use \(g = 9.8 m/s^2\))

• A. 23.6 N
• B. 94.4 N
• C. 59 N
• D. 35.4 N

Answer 1

The Correct Option is B

Force on mass \( m_1 \):

\( T_1 - T_2 - m_1 g = m_1 a \)

Substitute the given values:

\( T_1 - T_2 - 5 \times 9.8 = 5a \) 

Simplifying:

\( T_1 - T_2 = 59.0 \, \text{N} \)

Force on mass \( m_2 \):

\( T_2 - m_2 g = m_2 a \)

Substitute the given values:

\( T_2 = 3 \times (9.8 + 2) \)

Simplifying:

\( T_2 = 3 \times 11.8 = 35.4 \, \text{N} \)

Now, solving for \( T_1 \):

\( T_1 = T_2 + 59.0 = 35.4 + 59.0 = 94.4 \, \text{N} \)

Therefore, the force on mass \( m_1 \) is: \( T_1 = 94.4 \, \text{N} \).