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. 188 N
B. 2=368 N
C. 288 N
D. 168 N

Answer

D

Explanation

` m_1 g sin 30^@ -T =m_1 a`
` T - m_2 g sin 30^@ = m_2 a `
Adding (I) and (ii ) we , get
` a=(m_1 g sin 30^@ - m_2 g sin 30^@ )/( m_1 +m_2)`
` +((40-30 ) g//2) /((40+30 )) = (10 )/( 70 ) xx (9.8 )/(2) = 0.7 ms^(-2)`
From eqn (ii ) we get
` T= m_2 g sin 30^@ +m_2 a`
` =m_2 g sin 30^@ +(m_2) /(m_1 +m_2) ( m_1 g sin 30 ^@ - m_2 sin 30 ^@ )`
`=( 2 m_1 m_2 g sin 30^@ )/( m_1 +m_2) = ( 2 xx 40 xx 30 xx9.8 xx (1//2))/( 40+30 )`
`= ( 1200 )/( 70) xx9.8 = 168 N`