Laws Of Motion Question 432
Question: Two masses $ {m_1} $ and $ {m_2} $ are attached to the ends of a massless string which passes over a frictionless pulley attached to the top of an inclined plane. The angle of inclination of the plane is $ \theta $ . Take $ g=10m{{s}^{-2}} $ If$ m _{1}=10kg,m _{2}=5kg, $
$ \theta = 30^\circ, $ what is the acceleration of mass $ m_2 $ ?
Options:
A) 0
B) $ (2/3)m{{s}^{-2}} $
C) $ 5m{{s}^{-2}} $
D) $ 10/3m{{s}^{-2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ a=\frac{(m _{2}-m _{1}\sin \alpha )g}{(m _{1}+m _{2})}=\frac{(5-10\sin {{30}^{{}^\circ }})g}{(5+10)}=0 $
BETA
