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pratu043

  • 3 years ago

The speed of a cycle increases from 2.5 m/s to 5 m/s. The total mass of the cyclist including the cycle is 120 kg. Calculate the work done by the cyclist to increase the speed.

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  1. shamim
    • 3 years ago
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    i m trying to solve this problem

  2. shamim
    • 3 years ago
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    here u=2.5 v=5 m=120kg work done W=?

  3. shamim
    • 3 years ago
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    \[W=FS=maS\]

  4. shamim
    • 3 years ago
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    again we know\[v ^{2}=u ^{2}+2aS\]

  5. shamim
    • 3 years ago
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    \[or, 2aS=v ^{2}-u ^{2}\]

  6. shamim
    • 3 years ago
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    \[or, aS=\frac{ v ^{2}-u ^{2} }{ 2 }\]

  7. shamim
    • 3 years ago
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    putting this value of aS in the 1st equation we get

  8. shamim
    • 3 years ago
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    \[W=m(\frac{ v ^{2}-u ^{2} }{ 2 })\]

  9. shamim
    • 3 years ago
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    \[or, W=120 \times (\frac{ 5^{2}-2.5^{2} }{ 2 })\]

  10. shamim
    • 3 years ago
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    or, W=1125 joule

  11. shamim
    • 3 years ago
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    r u following me

  12. Vincent-Lyon.Fr
    • 3 years ago
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    Use work-energy theorem, and write: \(W=m(\Large\frac{ v ^{2}-u ^{2} }{ 2 })\) directly. This is more general than the derivation provided by shamim, as it will be true even for non-uniform acceleration. But this will only be true if the cyclist is on a horizontal road.

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