two masses 5kg and 2kg connected by a rope .The 5kg mass experiences a kinetic friction of 9.8N between it and the table. Determine the i) acceleration of the system ii) tension in the rope
Given:
"m_1=5\\:\\rm kg"
"m_2=2\\:\\rm kg"
"F_f=9.8\\:\\rm N"
The Newton's second law says
"m_1a=T-F_f\\\\\nm_2a=m_2g-T"So,
(i)
"(m_1+m_2)a=m_2g-F_f\\\\\na=\\frac{m_2g-F_f}{m_1+m_2}""a=\\frac{2*9.8-9.8}{5+2}=1.4\\:\\rm m\/s^2"
(ii)
"T=m_2g-m_2a=2*(9.8-1.4)=16.8\\:\\rm N"
Comments
Leave a comment