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Work Energy and Power question

2012 · Q157
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Work Energy and Power question

2012 · Q157

NEETPhysicsWork Energy and PowerMCQ+4 / −1
A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 m s−-−1. It collides with a horizontal spring of force constant 200 N m−-−1. The maximum compression produced in the spring will be
  1. A
    0.5 m
  2. B
    0.6 m
  3. C
    0.7 m
  4. D
    0.2 m
View written solutionFree

Correct answer: B

At maximum compression the solid cylinder will stop so loss in K.E. of cylinder = gain in P.E. of spring

⇒12mv2+12Iω2=12kx2 \Rightarrow {1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2} = {1 \over 2}k{x^2}⇒21​mv2+21​Iω2=21​kx2

⇒12mv2+12mR22(vR)2=12kx2 \Rightarrow {1 \over 2}m{v^2} + {1 \over 2}{{m{R^2}} \over 2}{\left( {{v \over R}} \right)^2} = {1 \over 2}k{x^2}⇒21​mv2+21​2mR2​(Rv​)2=21​kx2

⇒34mv2=12kx2 \Rightarrow {3 \over 4}m{v^2} = {1 \over 2}k{x^2}⇒43​mv2=21​kx2

⇒34×3×(4)2=12×200x2 \Rightarrow {3 \over 4} \times 3 \times {\left( 4 \right)^2} = {1 \over 2} \times 200{x^2}⇒43​×3×(4)2=21​×200x2

⇒36100=x2⇒x=0.6 m \Rightarrow {{36} \over {100}} = {x^2} \Rightarrow x = 0.6\,m⇒10036​=x2⇒x=0.6m

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