Question #201706

The system of blocks shown is initially moving with a velocity of 2m/sec. If they acquire a final velocity of 6m/sec after travelling 4m, find weight of block B. Use Work Energy principle.

Expert's answer

"W=\\triangle KE= \\frac{1}{2} mV^2f-\\frac{1}{2}mV^2i=\\frac{1}{2}m_{AB}*6^2-\\frac{1}{2}m_{AB}*2^2=18m_{AB}-2m_{AB}"

"=16m_{AB}=16(m_a+m_B)"

"W=Fd=4*9.8 (m_A+m_B)"

"16m_A+16m_B=39.24m_A+39.24m_B"

"16m_A-39.24m_A=39.24m_B-16m_B"

"-23.24m_A=55.24M_B"

"\\frac{m_A}{m_B}=\\frac{55.24}{23.24}"

"=m_B=23.24"

"W=mg=23.24*9.81=2227.9844N"

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