question_answer

Two blocks of masses \[{{m}_{1}}=10\,kg\,\,and\,\,{{m}_{2}}=20\,\,kg\] 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             \[\left[ Take g = 10 m/{{s}^{2}} \right]\]

A) \[\mu {{m}_{1}}g+\frac{\mu {{m}_{2}}g}{2}\]

B) \[\mu {{m}_{1}}g+\mu {{m}_{2}}g\]

C) \[\mu {{m}_{1}}g+\frac{\mu {{m}_{2}}g}{2}\]

D) \[\frac{\mu {{m}_{1}}g+\mu {{m}_{2}}g}{2}\]

Correct Answer: A

Solution :

[a] \[{{W}_{F}}+{{W}_{sp}}+{{W}_{Fric}}=\Delta k\] \[\Rightarrow  Fx-\frac{1}{2}k{{x}^{2}}-\mu mg\,\,x=0 and\,\,kx=\mu {{m}_{2}}g\] \[\Rightarrow  F-\frac{1}{2}\mu {{m}_{2}}g-\mu {{m}_{1}}g=0\Rightarrow  F=\mu {{m}_{1}}g+\frac{\mu {{m}_{2}}g}{2}\]