Two blocks of masses of 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure.

Options

A. `0.7 ms^-2`
B. `0.8 ms^-2`
C. `0.6 ms^-2`
D. `0.5 ms^-2`

Answer

A

Explanation

Here `m_1=40kg m_2=30 kg theta=30^@`

Let T be the tension in the string and a be the acceleration of the system of two masses.
Their equations of motion are
`m_1a=m_1g sin 30^@-T`……..(i)
`m_2a=T-m_@g sin 30^@`………(ii)
Adding (i) and (ii) we get
`(m_1+m_2)a=(m_1+m_2)g sin 30^@`
Substituting the values we get
`(40+30)a=(40-30) times 9.8 times 1/2=49`
or `a=49/70=0.7ms^-2`