Work Energy And Power Question 253
Question: Two blocks of masses $ m_1=10kgandm_2=20kg $ are connected by a spring of stiffness $ k=200N/m $ . The coefficient of friction between the blocks ’ and the fixed horizontal surface is $ \mu = 0.1 $ . Find the minimum constant horizontal force F (in newtons) to be applied to m, in order to slide the mass my $ [ Take g = 10 m/s^{2} ] $
Options:
A) $ \mu m_1g+\frac{\mu m_2g}{2} $
B) $ \mu m_1g+\mu m_2g $
C) $ \mu m_1g+\frac{\mu m_2g}{2} $
D) $ \frac{\mu m_1g+\mu m_2g}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ W _{F}+W _{sp}+W _{Fric}=\Delta k $
$ \Rightarrow Fx-\frac{1}{2}kx^{2}-\mu mgx=0 andkx=\mu m_2g $
$ \Rightarrow F-\frac{1}{2}\mu m_2g-\mu m_1g=0\Rightarrow F=\mu m_1g+\frac{\mu m_2g}{2} $
