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ICSE Class 10 Physics 2014 Paper

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Question : 8 of 66
Marks: +1, -0
Calculate the change in the Kinetic energy of a moving body if its velocity is reduced to 13\frac{1}{3} rd of the initial velocity.
Solution:  
Given : Initial velocity =v;=v ; Final velocity =  v3=\;\frac{v}{3}
Initial kinetic energy k1=12×m×v2k_1 = \frac{1}{2} \times m \times v^2
Final kinetic energy k2=12×m×(v3)2k_2 = \frac{1}{2} \times m \times \left(\frac{v}{3}\right)^2
=19×12mv2= \frac{1}{9} \times \frac{1}{2} m v^2
Change in kinetic energy
  =12mv2−19×12mv2\;= \frac{1}{2} m v^2 - \frac{1}{9} \times \frac{1}{2} m v^2
  =12mv2(1−19)=89×(12mv2)\;= \frac{1}{2} m v^2 \left(1 - \frac{1}{9}\right) = \frac{8}{9} \times \left( \frac{1}{2} m v^2 \right)
=89×Initial kinetic energy.= \frac{8}{9} \times \text{Initial kinetic energy.}
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