Question

Find the magnitude and the direction of acceleration of the body $m_2$ when the formerly stationary system of masses starts moving.

• A. $5g$
• B. $0.05g$
• C. $0.005g$
• D. $0.5g$

Answer 1

Answer: (B)

$0.05g$

For mass $m_1:m_{1}a=T-m_{1}g\sin \alpha -k{m}_{1}g\cos \alpha .....\left(1\right)$
For mass $m_2:m_{2}a=m_{2}g-T....\left(2\right)$
solving (1) and (2), $m_{1}a=m_{2}g-m_{2}a-m_{1}g\sin \alpha -k{m}_{1}g\cos \alpha$
or,$(m_1+m_2)a=(m_2-m_1\sin \alpha -k{m}_1\cos \alpha )$
or, $a=g[\frac{m_2}{m_1+m_2}-\frac{m_1}{m_1+m_2}(\sin \alpha +k\cos \alpha \left)\right]$
or,$a=\frac{g}{1+\eta }[\eta -\sin \alpha -k\cos \alpha )]$
or,$a=\frac{g}{1+\frac{2}{3}}[\frac{2}{3}-\sin {30}^o-0.10\cos {30}^o]=0.05g$