Question

# Two blocks of masses $${m}_{1}$$ and $${m}_{2}$$ are connected by a spring of stiffness $$k$$. The coefficient of friction between the blocks and the surface is $$\mu$$. Find the minimum constant force $$F$$ to be applied to $${m}_{1}$$ in order to just slide the mass $${m}_{2}$$

Solution

## At any intermediate position, the net external force is equal to$$F'=F-\left( \mu { m }_{ 1 }g+kx \right)$$(numerically)The work done by the net external force $$F'$$ in displacing the block $$m$$, through a length $${ x }_{ 0 }={ W }_{ ext }=\int { F' } .dx$$$$\Rightarrow { W }_{ ext }=\int _{ 0 }^{ { x }_{ 0 } }{ \left[ F-\left( \mu { m }_{ }g+kx \right) \right] dx }$$where $${x}_{0}=$$ compression of the spring required to just move the block $${m}_{2}$$$$\Rightarrow k{ x }_{ 0 }=\mu { m }_{ 2 }g\Rightarrow { x }_{ 0 }\cfrac { \mu { m }_{ 2 }g }{ k }$$Since the KE of the system just before moving the block $${m}_{1}$$, just before moving the block $${m}_{2}$$ iz zero, the change in $$KE=0$$$$\Rightarrow \Delta KE=0$$$$\therefore { W }_{ ext }=\Delta KE$$$$\Rightarrow \int _{ 0 }^{ { x }_{ 0 } }{ \left[ F-\left( \mu { m }_{ }g+kx \right) \right] dx } =0\Rightarrow F=\mu { m }_{ 1 }g+\cfrac { k{ x }_{ 0 } }{ 2 } =\mu { m }_{ 1 }g+\cfrac { \mu { m }_{ 2 }g }{ 2 } =\left[ { m }_{ 1 }+\cfrac { { m }_{ 2 } }{ 2 } \right] \mu g$$Physics

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