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Question

A block of mass m equals 25kg sliding on a smooth horizontal surface with a velocity v equals 3 m s to the power of negative 1 end exponent meets the spring of spring constant k equals 100 N m to the power of negative 1 end exponent fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as is returns to the original position respectively are

  1. 1.5 blank m comma blank minus 3 blank m s to the power of negative 1 end exponent    
  2. 1.5 blank m comma blank 0.01 blank m s to the power of negative 1 end exponent    
  3. 1.0 blank m comma blank 3 blank m s to the power of negative 1 end exponent    
  4. 0.5 blank m comma blank 2 blank m s to the power of negative 1 end exponent    

The correct answer is: 1.5 blank m comma blank minus 3 blank m s to the power of negative 1 end exponent


    When block strikes the spring, the kinetic energy of block converts into potential energy of spring ie,
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
    Or x equals square root of fraction numerator m v to the power of 2 end exponent over denominator k end fraction end root
    equals square root of fraction numerator 25 cross times 3 to the power of 2 end exponent over denominator 100 end fraction end root square root of fraction numerator 9 over denominator 4 end fraction end root equals fraction numerator 3 over denominator 2 end fraction equals 1.5 blank m
    When block returns to the original position, again potential energy converts into kinetic energy of the blocks, so velocity of the block is same as before but its sign changes as it goes to mean position.
    H e n c e v equals negative 3 m s to the power of negative 1 end exponent

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