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Physics

Mechanics

Easy

Question

Consider the situation as shown in the diagram. The pulley has radius R and moment of inertia I. The block is released when the spring (spring constant K) is unstretched. Find the speed of block when the block has fallen a distance h. Assume no slipping.

$\sqrt{\frac{2 m g h - K h^{2}}{m [1 + \frac{I}{m R^{2}}}}]$
$\sqrt{\frac{2 m g h}{K}}$
$\sqrt{\frac{K h^{2}}{m}}$
$\frac{\sqrt{\frac{m g h - K h^{2}}{I}}}{1 + \frac{I}{m R^{2}}}$

The correct answer is: $open square root of fraction numerator 2 m g h minus K h squared over denominator m open square brackets 1 plus fraction numerator I over denominator m R squared end fraction close end fraction end root close square brackets$

$table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell m g h equals 1 half m v squared plus 1 half I omega squared plus 1 half K h squared semicolon space of 1em omega equals v over R end cell row cell m g h minus 1 half K h squared equals 1 half m v squared plus 1 half I open parentheses v over R close parentheses squared equals 1 half m v squared open parentheses 1 plus fraction numerator I over denominator m R squared end fraction close parentheses end cell row cell v equals square root of open fraction numerator 2 m g h minus K h squared over denominator m open parentheses 1 plus fraction numerator I over denominator m R squared end fraction close end fraction close parentheses end root end cell end table$

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