Answer to Question #201706 in Civil and Environmental Engineering for Akshat Garg

Question

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.

Answer 1

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

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

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

$16m_A+16m_B=39.24m_A+39.24m_B$

$16m_A-39.24m_A=39.24m_B-16m_B$

$-23.24m_A=55.24M_B$

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

$=m_B=23.24$

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