Physics-
General
Easy

Question

The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k equals a s to the power of 2 end exponent where a is a constant

  1. 2 a fraction numerator s to the power of 2 end exponent over denominator R end fraction  
  2. 2 a s open parentheses 1 plus fraction numerator s to the power of 2 end exponent over denominator R to the power of 2 end exponent end fraction close parentheses to the power of 1 divided by 2 end exponent  
  3. 2 a s  
  4. 2 a fraction numerator R to the power of 2 end exponent over denominator s end fraction  

The correct answer is: 2 a s open parentheses 1 plus fraction numerator s to the power of 2 end exponent over denominator R to the power of 2 end exponent end fraction close parentheses to the power of 1 divided by 2 end exponent


    According to given problem fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals s to the power of 2 end exponent
    rightwards double arrow v equals s square root of fraction numerator 2 a over denominator m end fraction end root
    S o a subscript R end subscript equals fraction numerator v to the power of 2 end exponent over denominator R end fraction equals fraction numerator 2 a s to the power of 2 end exponent over denominator m R end fraction (i)
    Further more as
    a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals fraction numerator d v over denominator d s end fraction. fraction numerator d s over denominator d t end fraction equals v fraction numerator d v over denominator d s end fraction(ii)
    Which is light of equation (i) i.e. v equals s square root of fraction numerator 2 a over denominator m end fraction end root yields
    a subscript t end subscript equals open square brackets s square root of fraction numerator 2 a over denominator m end fraction end root close square brackets open square brackets square root of fraction numerator 2 a over denominator m end fraction end root close square brackets equals fraction numerator 2 a s over denominator m end fraction(iii)
    So that a equals square root of a subscript R end subscript superscript 2 end superscript plus a subscript t end subscript superscript 2 end superscript end root equals square root of open square brackets fraction numerator 2 a s to the power of 2 end exponent over denominator m R end fraction close square brackets to the power of 2 end exponent open square brackets fraction numerator 2 a s over denominator m end fraction close square brackets to the power of 2 end exponent end root
    Hence a equals fraction numerator 2 a s over denominator m end fraction square root of 1 plus open square brackets s divided by R close square brackets to the power of 2 end exponent end root
    therefore F equals m a equals 2 a s square root of 1 plus open square brackets s divided by R close square brackets to the power of 2 end exponent end root

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