Question #79461
Two blocks of 100 and 50 kg are connected by a cord passing over a frictionless pulley and resting on frictionless plane of angle 37 and 53 respectively find acceleration
Expert's answer
Newton’s second law gives
m_1 a=m_1 g sin〖53°〗-T
m_2 a=-m_2 g sin〖37°〗+T
Where T is tension in a cord
Thus
(m_1 〖+m〗_2 )a=g(m_1 sin〖53°〗-m_2 sin〖37°〗 )
a=g (m_1 sin〖53°〗-m_2 sin〖37°〗)/(m_1 〖+m〗_2 )
a=9.81 (100 sin〖53°〗-50 sin〖37°〗)/(100+50)=3.25 m/s^2
m_1 a=m_1 g sin〖53°〗-T
m_2 a=-m_2 g sin〖37°〗+T
Where T is tension in a cord
Thus
(m_1 〖+m〗_2 )a=g(m_1 sin〖53°〗-m_2 sin〖37°〗 )
a=g (m_1 sin〖53°〗-m_2 sin〖37°〗)/(m_1 〖+m〗_2 )
a=9.81 (100 sin〖53°〗-50 sin〖37°〗)/(100+50)=3.25 m/s^2
